Post on 07-Apr-2015
XtraEdge for IIT-JEE 1 APRIL 2010
Dear Students,
Motivate Yourself One of the greatest virtues of human beings is their ability to think and act accordingly. The emergence of the techno savvy human from the tree swinging ape has really been a long journey. This transition has taken a span of countless centuries and lots of thinking caps have been involved. Inquisitiveness and aspiration to come out with the best have been the pillars for man's quest for development. Self-motivation is the sheer force, which pulled him apart and distinguished him from his primitive ancestors. Many times, in our life, when we are reviving old memories we get into a phase of nostalgia. We feel that we could have done better than what we had achieved. But thinking back won’t rewind the tireless worker called time. All we can do is promise ourselves that we will give our very best in the future. But do we really keep up to our mental commitments? I can guess that 90% answers are in the negative. This is because of that creepy careless attitude which is slowly, but surely entering into our mind. We easily forget the pains of yesterday to relish the joys of today. This is the only time in our life, when we can control our fate, by controlling our mind. So it is time to pull up our socks and really motivate ourselves so that we can give our best shot in the future. Self-motivation is the need of the hour. Only we can control and restrict ourselves. It’s up to us, how we use our mental capabilities to the best of our abilities. Here are some Funda's for self-motivation. Don't just read them digest each one of them and apply them and I bet it will make a better YOU. • The ultimate motivator is defeat. Once you are defeated, you
have nowhere to go except the top. • Then only thing stopping you is yourself. • There is no guarantee that tomorrow will come. So do it today. • Intentions don't count, but action's do. • Don't let who you are, stunt what you want to be. • Success is the greatest motivator. • Your goals must be clear, but the guidelines must be flexible. Try to include these one liners in your scrapbook or on your favorite poster. You will be sub-consciously tuned to achieve what you want. Also do keep in mind that nothing can control your destiny but you! With Best Wishes for Your Future.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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Editor : Pramod Maheshwari
To fly, we have to have resistance Volume - 5 Issue - 10
April, 2010 (Monthly Magazine)
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XtraEdge for IIT-JEE 2 APRIL 2010
XtraEdge for IIT-JEE 3 APRIL 2010
Volume-5 Issue-10 April, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Key Concepts & Problem Solving strategy for IIT-JEE.
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics, Chemistry & Maths
Much more IIT-JEE News.
Xtra Edge Test Series for JEE – 2010 & 2011
Success Tips for the Month
• "The way to succeed is to double your error rate."
• "Success is the ability to go from failure to failure without losing your enthusiasm."
• "Success is the maximum utilization of the ability that you have."
• We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.
• Along with success comes a reputation for wisdom.
• They can, because they think they can.
• Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.
• Keep steadily before you the fact that all true success depends at last upon yourself.
CONTENTS
INDEX PAGE
NEWS ARTICLE 4 • President buries 'Time Capsule' on IIT Kanpur campus • IITs admission criteria set for an overhaul
IITian ON THE PATH OF SUCCESS 6 Mr. Krishnamurthy Rengarajan
KNOW IIT-JEE 7 Previous IIT-JEE Question
XTRAEDGE TEST SERIES 52 Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE Mock Test BIT SAT SOLUTIONS 92
Regulars ..........
DYNAMIC PHYSICS 14
8-Challenging Problems [Set# 12] Students’ Forum Physics Fundamentals Calorimetry, K.T.G., Heat Transfer Atomic Structure, X-Ray & Radio Activity CATALYST CHEMISTRY 30
Key Concept Aromatic Hydrocarbon Solubility Product Understanding: Organic Chemistry
DICEY MATHS 38
Mathematical Challenges Students’ Forum Key Concept Calculus Algebra
Study Time........
Test Time ..........
XtraEdge for IIT-JEE 4 APRIL 2010
President buries 'Time Capsule' on IIT Kanpur campus
Kanpur: President Pratibha Patil recently buried a 'Time Capsule' on the Indian Institute of Technology, Kanpur (IIT-K) campus on the occasion of its golden jubilee celebrations and also unveiled a nanosatellite developed by the institute. The capsule, which is made of a special metal, contains pen-drives, chips, images and several other documents related to the landmark achieve-ments of the IIT-K. Lauding the nanosatellite Jugnu's development team, Patil said it projects the complex nature of tasks that the students there were equipped to handle. Congratulating IIT-K students and faculty, Patil said that the institute has come a long way in its 50 years of its existence, and also called upon the institute's students and faculty members to develop such devices that can harness energy in efficient ways with minimal negative impact on the environment. "It (IIT-K) had made an impact on technical education within the country, while its students through their innovations, have played an important role in India, as well as around the world," the President said. Jugnu, developed by a team of 50 IIT students, will help in collection of information related to floods, drought and other natural calamities.
IITs admission criteria set for an overhaul New Delhi: The admission criteria for admission to the premier engineering institutes of the country - the Indian Institutes of Technology (IITs), is all set for an overhaul.
According to the recommendation of a panel set up under IIT Kaharagpur's Director, Mr. Damodar Acharya, the Joint Entrance Examination (JEE) for undergraduates and Graduate Aptitude Test in Engineering (GATE) for postgraduates may soon not be the only criteria for their admission. The committee proposes to consider the Class XII marks as well.
Presently, the eligibility for a student to sit for a JEE is 60% in class XII after which a cut-off is decided every year for admission into an IIT. Following the consensus of the last meeting of the IIT Council to give more weightage to the school leaving examination, the committee proposes to mull over the issue of factoring in class XII examination result in the cut-off for admission to the IITs.
The IIT Directors in a meeting with the Human Resource Development (HRD) Minister, Kapil Sibal, in Manesar, also expressed their discontent over the existing pattern for GATE as they felt that not enough students with research orientation were being picked up through this exam.
Sibal said that each IIT should submit a proposal within a month on one area of expertise in higher
research. Although the admission criteria is becoming stringent with the proposal of IIT-Kharagpur Director's recommendation, the HRD minister retreated with a firm 'No' to IIT-Kanpur Director's proposal for a fee hike. According to the fee-hike proposal, which the IIT-Kanpur's Director framed, it suggested to hike fees by eight times. At present, the B.Tech students pay Rs.50,000 per year as their fee which the committee proposed to increase up to Rs.4 lakh per annum over the period of 10 years with a Rs.35,000 mark-up every year.
"The proposal on fee hike should be discussed as the government was planning to set up a Higher Education Funding Corporation which would address poor students," said a senior official.
During the meeting, Sibal also asked the IITs to come up with their plans for the future endeavors in a specialized area in which they want to emerge as global giants by 2020 in four weeks time.
Nanotech is used to treat cancer - IIT-B & Docs In Mumbai it could have been India’s second nano success if it passes the muster. Only this nano creation is revealed in the healthcare field and it was possible because of partnership between oncologists and scientists of the Indian Institute of Technology-Bombay.
The joint effort of the IIT -B and doctors from Tata Memorial
XtraEdge for IIT-JEE 5 APRIL 2010
Hospital in Parel and Apollo Hospital in Hyderabad has the possibility to transform treatment of retinoblastoma—a rare cancer of the retina that mainly affects children under two years of age. They have developed a nano-particle that could conquer the child killer.
Shirin Thakur, Guntur-based teenager was suffering from recurrent retinoblastoma since she was two years old. Last week, she took the third injection of a special mixture into the tissues around her left eye which was made of nano-particles of carboplatin which is commonly used to treat retinoblastoma. While she was standing with her doctor in the Apollo hospital, Hyderabad said, “I have been suffering from attacks of retinoblastoma in my left eye since I was two. Even in the US, they told me there is no hope but to remove my eye.’’ Now, she has “fuzzy’’ vision in the nearly blind eye. “My vision gets better every day.’’
PM was asked to fine-tune JEE by Maths Wizard PM Manmohan singh was given suggestions on the improvement of IIT-JEE pattern by Maths Wizard Anand Kumar. He pleaded that poor aspirants should be given at least three attempts for IIT-JEE exam, as they are usually the late starters.
Every year Anand’s Super-30 offers free accommodation to 30 poor students and provides coaching to them free of cost in order to help them to crack JEE. This noble initiative, has of late reported 100% success rate with all its 30 students making it to IIT’S, without seeking any financial support from government or non-government sources.
In his 15-minute interaction with the PM in New Delhi, Anand enlightened him of his efforts usher in a new awakening in Bihar by sending 182 poor students to IITs within the past seven years. He further informed the PM that he has decided to increase the student intake to 60 from 30. Very soon he would open schools for talented poor children so as to provide them the right momentum at the secondary school level itself. He added that government should run coaching programmes on the pattern of Super-30 for talented students from the countryside and it should not be confined to IIT only.
Audibly elated Anand told TOI over the phone that Pm patted him on his back and also instructed PMO to look into his suggestions.
He said that many problems asked in JEE are of Maths Olympiad level so students from villages find it difficult to solve such problems even if they have sound knowledge of the subject at +2 level. He pleaded that the exam should be designed in such a way that conceptual or analytical problems of the +2 level should be asked in JEE."
IIT Rajasthan moves to state varsity campus Jaipur: The Indian Institute of Technology, Rajasthan (IIT-R) has been temporarily shifted to the engineering faculty building of Jai Narain Vyas University according to a Memorandum of Understanding (MoU) signed in Jodhpur over the weekend. Students of IIT-R earlier had to rely on IIT-Kanpur and other institutions to attend classes.
IIT-Rajasthan founder-director P. K. Kalra and JNV University Registrar Nirmala Meena, in the
presence of Chief Minister Ashok Gehlot who was on a visit to Jodhpur, signed the MoU.
The CM also inaugurated the work to extend the building. Gehlot pointed out in an official press release that the establishment of IIT in Jodhpur was recommended and headed by the noted economist V.S. Vyas after he studied the facilities and educational standard of various towns in the State.
"The final decision to nod for an IIT in the city was accepted at the central level," added Gehlot. He further announced that the construction of a new building in Rajasthan would require some time, and for the time being the classes would be held in JNV University while those in the extended portion of the building would begin from May his year.
Following special efforts made by the state government Mr. Gehlot said that the union government has also nodded to use the newly constructed A.S.K. Hostel in the university for IIT students.
"The establishment of an IIT in the state would definitely boost its higher education and will prove to be a milestone," said Gehlot.
According to the MoU, it approves the functioning of IIT-Rajasthan only on temporary basis for two years in the engineering faculty building.
State Technical Education Minister Mahendrajit Singh Malviya and JNV University Vice-Chancellor Naveen Mathur signed the MoU as witnesses. Among others, Jodhpur MP Chandresh Kumari, Barmer-Jaisalmer MP Harish Chaudhary, Jodhpur Mayor Rameshwar Dadich, Deputy Mayor Niyaz Mohammed and Collector Naveen Mahajan were present on the occasion.
XtraEdge for IIT-JEE 6 APRIL 2010
Knowledge is indeed wealth. Who better exemplifies it than Krishnamurthy Rengarajan,IIT-B gold medallist (B Tech dual-degree course). Krishnamurthy's story is that of hard work, sheer grit and determination. His undying passion for learning and excellence has paid off. Coming from a lower middle class background, Krishnamurthy has made his parents proud when he passed with flying colours. His father, who works as a typist at Bharatiya Vidya Bhavan is overwhelmed by his son's achievement. Rengarajan, who hails from Tamil Nadu, came to Mumbai 28 years ago and settled down in a distant Mumbai suburb of Dombivli. Though the family went through a lot of hardships initially, he made sure that his children were well educated. "My son always wanted to join the IIT. When people asked him what if you don't get through the entrance examinations, he used to say, `there is no question of me not clearing the test," says his proud father. And, of course, he did top all the five years at IIT, a result of sheer hard work and brilliance, says his mother, barely able to control her excitement. "I am very happy for him," says Radha Rengarajan. Krishnamurthy did his schooling at the Kidland School in Dombivli and pre-degree from V G Vaze College at Mulund. His favourite subject being mathematics it was obvious that he would pursue a degree in engineering. He won the Rakesh Mathur award of Rs 1 lakh (Rs 100,000) during his third year and other scholarships throughout the four years. Here's what Krishnamurthy had to say on his IIT experience. My IIT experience The five years I spent at IIT were the best in my life. I will cherish each and every moment here. I loved everything here: the professors are the best one can ever get, the facilities to study and the extra-curricular activities are excellent. I made best of friends and thoroughly enjoyed my college life. I don't think I will ever get this experience anywhere else.
On studies Before joining IIT, I used to study for 7 to 8 hours daily. After joining IIT, I used to spend about a couple of hours. I did not go for anything coaching classes. I learnt through Brilliant Tutorial correspondence course and my preparation began after I finished my 10th standard. Why IIT IIT is one of the premier institutes in India. I always wanted to get good higher education, so I opted for IIT. My mantra for success There is no short cut to success. One has to work very hard, put in a lot of effort, should have a problem-solving mentality and a right approach to every problem. My parents always stood by me, their support has been invaluable and am overwhelmed. Advice to IIT aspirants Work hard. You have to spend a lot of time preparing as exams are getting more and more competitive. You must also have problem-solving skills. Interests Solving math puzzles, reading books. I used to play cricket, but now I don't get the time. Next move Money is the least important thing for Krishnamurthy. So no jobs for the time being. "I have been selected for the scholarship programme at Stanford University for a PhD in operations research. I would like to research on optimising computer networks and operation systems. Quality research is available abroad. After the PhD programme I would like to join any academia of good repute and continue my research activities. Among corporates, I admire Google. It is the one company that reflects perfection, hard work and efficiency." Will you come back to India? Of course, I will. The brain drain phenomenon is dying out. It's the time for reverse brain.
Mr. Krishnamurthy Rengarajan IIT-B Gold Medallist
Success StoryThis article contains story of a person who is successful after graduation from different IIT's
XtraEdge for IIT-JEE 7 APRIL 2010
PHYSICS
1. A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure in solution). The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surface are frictionless. The ball is given a gentle push (towards the right see figure in solution) The angle made by the radius vector of the ball with the upward vertical is denoted by θ [IIT-2002]
(a) Express the total normal reaction force exerted by the sphere on the ball as a function of angle θ.
(b) Let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the sphere A and B, respectively. Sketch the variations of NA and NB as functions of cos θ in the range 0 ≤ θ ≤ π by drawing two separate graphs in your answer book, taking cos θ on the horizontal axes.
Sol. The ball is moving in a circular motion. The necessary centripetal force is provided by
(mg cos θ – N). Therefore
V mgsinθ
mgcosθ mg
R θ θ
C d/2 NA
D A
B
mg cos θ – NA =
+
2dR
mv2 …(i)
According to energy conservation
21 mv2 = mg
+
2dR (1 – cos θ) …(ii)
From (i) and (ii) NA = mg (3 cos θ – 2) …(iii) The above equation shows that as θ increases NA
decreases. At a particular value of θ, NA will become zero and the ball will lose contact with sphere A. This condition can be found by putting NA = 0 in eq. (iii) 0 = mg (3 cos θ – 2)
∴ θ = cos–1
32
The graph between NA and cos θ From eq (iii) when θ = 0, NA = mg. 2mg
NA
mg
cos θ 1
2mg
cos θ–1
5mg
NB
When θ = cos–1
32 ; NA = 0
The graph is a straight line as shown.
When θ > cos–1
32
NB – (– mg cos θ) =
2dR
mv2
+
⇒ NB + mg cos θ =
+
2dR
mv2 …(iv)
Using energy conservation
2mv21 = mg
θ
+−
+ cos
2dR
2dR
+
2dR
mv2 = 2 mg [1 – cos θ] …(v)
From (iv) and (v) we get NB + mg cos θ = 2 mg – 2mg cos θ NB – mg (2 – 3 cos θ)
When cos θ = 32 , NB = 0
When cos θ = – 1, NB = 5 mg Therefore the graph is as shown.
2. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed coaxially on a a thin metal disc of mass 0.4 Kg and the same cross section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the materical of all cylinder is 10 watt/kg. K, how long will it take for the temperature of the disc to increases to 350 K? Assume, for purpose of calculation, the thermal conductivity of the disc to be very high and
KNOW IIT-JEE By Previous Exam Questions
XtraEdge for IIT-JEE 8 APRIL 2010
the system to be thermally insulated except for the upper face of the cylinder. [IIT-1992]
Sol. Initially the temperature at the open end is 400 K and the temperature at the metal disc-cylinder interface is 300 K.
Metal disc Insulation
Cylinder
H
Openend
θ
As heat passes through the cylinder and reaches the
metal disc, the temperature of metal disc rises. Since the conductivity (thermal) of metal disc is very high so temperature of the whole disc will rise and along with that the temperature of the other end of the cylinder (metal disc-cylinder interface) also rises simultaneously. Let at any instant of time, the temperature of the metal disc-cylinder interface is θ. At this instant the rate of heat crossing the cylinder.
dtdQ =
l
)–400(KA θ ...(i)
where K = thermal conductivity A = area of cross section of cylinder l = length of cylinder The same amount of heat is received by the metal
disc. Therefore
dtdQ = mc
dtdθ ...(ii)
m = mass of disc c = specified heat of metal disc From (i) and (ii)
mcdtdθ =
l
)–400(KA θ
θ−θ
400d ×
KAmcl = dt
On integrating At t = 0, θ = 300
∫t
0dt =
KAmcl
∫ θ−θ350
300 )400(d At t = t, θ = 350
⇒ t = KAmcl− [log (400 – θ) 350
300]
= 04.010
303.24.06004.0×
×××− log10 300400350400
−−
= 166.38 sec.
3. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields E and B are present. E and B are parallel to each other. At time t = 0 the velocity v0 of the particle is perpendicular to E. (Assume that its speed is always <<c, the speed of light in vaccum.) Find the velocity v of the particle at time t. You must express your answer in terms of t, q, m, the vectors v0, E and B and their magnitudes v0, E and B. [IIT-1998]
Sol. Because the forces due to parallel electric and magnetic fields on a charged particle moving perpendicular to the fields will be at right angles to each other (electric force being along the direction of →E while magnetic force perpendicular to the plane containing v and B ) so magnetic force will not affect the motion of charged particle in the direction of electric field and vice-versa. So the problem is equivalent to superposition of two independent motions as shown in the adjoining figures.
Y
X
Zv0 Fm
k
j
i
B E Fe
So for motion of the particle under electric field
alone,
ay = mqE i.e.,
dtdvy =
mqE
or ∫y
0ydv = ∫
y
0 mqE dt i.e., vy =
mqE t ...(1)
While at the same instant, the charged particle under the action of magnetic field will describe a circle in the x-z plane with
r = qB
mv0 i.e., ω = r
v0 = mqB
Xv0
v0cos θ
v0
θ
θ
Z
v0 sin θ
So angular position of the particle at time t in the x-z
plane will be given by
θ = ωt = mqB t
and therefore in accordance with figure
vx = v0 cosθ = v0 cos ωt = v0 cos
t
mqB ...(2)
and vz = v0 sinθ = v0 sin ωt = v0sin
t
mqB ...(3)
So in the light of equations (1), (2) and (3), we get
→v = i vx + j vy + k vz
= i v0 cos
t
mqB + j
tmqE + v0 sin
tmqB
k
XtraEdge for IIT-JEE 9 APRIL 2010
But because here, i = 0
0
vv→
, j = EE→
= BB→
and k = BvBv
0
0 ×
So, →0v =
→
0
0
vv
v0 cos
mqBt +
→
EE
mqE t
+ BvBv
0
0 × v0sin
mqBt
4. A long solenoid of radius 'a' and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d<<R) and length L. A variable current i = i0 sin ωt flows through the coil. If the resitivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT-2005]
Sol. The magnetic field in the solenoid is given by B = µ0ni
a
L
d R
⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given] The magnetic flux linked with the solenoid
φ = →B .
→A = BA cos 90º
= (µ0 n i0 sin ωt) (πa2) ∴ The rate of change of magnetic flux through the
solenoid
dtdφ = π µ0 n a2 i0 ω cos ωt
The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagetic induction, the induced emf produced in the cylinderical shell is
ITOP VIEW
e = –dtdφ = – πµ0 n a2 i0 ω cos ωt ...(i)
The resistance offered by the cylindrical shell to the flow of induced current I will be
R = ρAl
Here, l = 2πR,
A = L × d
∴ R = ρLd
R2π ...(ii)
The induced current I will be
I = R
|e| = R2
Ld]tcosinaµ[ 02
0
π×ρ×ωωπ
I = R2
Ldtcosinaµ 02
0
ρ×ωωπ
5. A beam of light has three wavelength 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 × 10–3 Wm–2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photo electrons liberated in two seconds. [IIT-1989]
Sol. Work function = 2.3 eV = 2.3 × 1.6 × 10–19J = 3.68 × 10–19 J Energy of a photon of wave length (4144 Å)
= λhc = 10
834
104144103106.6
−
−
××××
= 4.78 × 10–3 × 10–16 = 4.78 × 10–19 J Energy of a photon of wave length 4972 Å
= 10
834
104972103106.6
−
−
×××× = 3.98 × 10–19J
Energy of a photon of wavelength 6216 Å
= 10
834
106216103106.6
−
−
×××× = 3.18 × 10–19 J
This means that light of wavelength 4144 Å and 4972Å are capable of ejection of electrons from the metal surface
The energy incident on 1 cm2 area of metal per second = 3.6 × 10–3 × 10–4 = 3.6 × 10–7 J
The energy/s of wavelength 4972 Å = 1.2 × 10–7 J No of photon incident of wavelength 4972 Å
E = λhcn ∴ n =
hcEλ
⇒ n = 834
107
103106.6104972102.1
××××××
−
−−
= 301.33 × 109
= 3.01 × 1011 Similarly number of photon incident of wavelength
4144 Å.
n = 834
107
103106.6104144102.1
××××××
−
−−
= 2.51 × 1011
⇒ Total number of photons capable of ejection of electrons per second
= 3.01 × 1011 + 2.51 × 1011 = 5.52 × 1011 ∴ Total number of photoelectrons ejected in two
seconds ≈ 11 × 1011.
XtraEdge for IIT-JEE 10 APRIL 2010
CHEMISTRY
6. From the following data, form the reaction between A and B. [IIT-1994]
Initial rate (mol L–1s–1) [A]
mol L–1 [B]
mol L–1 300 K 320 K 2.5 ×10–4 3.0 ×10–5 5.0 ×10–4 2.0 × 10–3
5.0 × 10–4 6.0 × 10–4 4.0 × 10–3 – 1.0 × 10–3 6.0 × 10–5 1.6 × 10–2 –
Calculate (a) the order of reaction with respect to A and with
respect to B, (b) the rate constant at 300 K, (c) the energy of activation, (d) the pre exponential factor.
Sol. Rate of reaction = k[A]l [B]m where l and m are the order of reaction with respect
to A and B respectively. From the given data, we obtain following expressions :
5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m ...(i) 4.0 × 10–3 = k[5.0 × 10–4] l [6.0 × 10–5]m ...(ii) 1.6 × 10–2 = k[1.0 × 10–3]l [6.0 × 10–5]m ..(iii) From eq. (ii) and eq. (iii), we get
2
3
106.1100.4
−
−
×× =
l
××
−
−
3
4
100.1100.5
or 0.25 = (0.5)l or (0.5)2 = (0.5) l or l = 2 From eq. (i) and eq. (ii), we get
3
4
100.1100.5
−
−
×× =
m
5
52
4
4
100.6100.3
100.5105.2
××
××
−
−
−
−
or 81 =
41 ×
m
21
or 21 =
m
21
or m = 1 (b) At T1 = 300 K,
k1 = 12 ]B[]A[reactionofRate =
]100.3[]105.2[100.5
524
4
−−
−
×××
= 2.67 × 108 L2 mol–2 s–1 (c) At T2 = 320 K,
k2 = 12 ]B[]A[reactionofRate
= ]100.3[]105.2[
100.2524
3
−−
−
×××
= 1.067 × 109 L2 mol–2 s–1
We know, 2.303 log 1
2
kk =
−
21
12a
TTTT
RE
or 2.303 log 8
9
1067.210067.1
×× =
×−
300320300320
314.8Ea
or 2.303 × 0.6017 =
×30032020
314.8Ea
or Ea = 20
300320314.86017.0303.2 ××××
= 55.3 kJ mol–1 (d) According to Arrhenius equation, k = RT/EaAe−
or 2.303 log k = 2.303 log A – RTEa
At 300 K,
2.303 log (2.67 × 108) = 2.303 log A – 300314.8
103.55 3
××
or 2.303 × 8.4265 = 2.303 log A – 22.17
or logA = 303.2
17.224062.19 + = 303.25762.41 = 18.0531
A = Antilog 18.0531 = 1.13 × 1018 s–1
7. 1 g of a mixture containing equal number of moles of carbonates of two alkali metals, required 44.4 ml of 0.5 N HCl for complete reaction. The atomic weight of one metal is 7, find the atomic weight of other metal. Also calculate the amount of sulphate formed on quantitative conversion of 1.0 g of the mixture in two sulphates. [IIT-1972]
Sol. Let, Mass of one alkali metal carbonate M2CO3 = xg Then, mass of other alkali metal carbonate M2´CO3
= (1 – x)g Step 1.
Equivalent mass of M2CO3 = 2
massMolecular = 2
74
= 37 (at mass of M = 7)
Meq. of M2CO3 = 37x ×1000
Moles of M2CO3 = 74x
Step 2
Equivalent mass of M2´CO3 = 2
massMolecular
= 2
60m2 +
( m = atomic mass of M´)
Meq. of M2´CO3 = 60m2
)x1(2+− × 1000
Moles of M2´CO3 = 60m2
x1+
−
Step 3. Meq. of HCl = NHCl × VHCl = 0.5 × 44.4 Step 4. According to the question, Moles of M2CO3 = Moles of M2´CO3
or 74x =
60m2x1
+− ...(i)
XtraEdge for IIT-JEE 11 APRIL 2010
And Meq. of M2CO3 + Meq . of M2´CO3 = Meq. of HCl
or 37x × 1000 +
60m2)x1(2
+− × 1000
= 0.5 × 44.4 ...(ii) Solving eq. (i) and (ii), we get m = 23 and x = 0.41 ∴ Mass of M2CO3 = x = 0.41 g and Mass of M2´CO3 = 1 – x = 0.59 g Step 5.
Equivalent mass of M2SO4 = 2
massMolecular
= 2
110 = 55
Meq. of M2SO4 = 55
W42SOM × 1000
But, Meq. of M2SO4 = Meq. of M2CO3
∴ 55
W42SOM × 1000 =
3741.0 × 1000
or 4SOM2W = 0.6095 g
Step 6.
Equivalent mass of M2´SO4 = 2
massMolecular
= 2
142 = 71
Meq. of M2´SO4 = 71
W42 SO´M × 1000
But, Meq. of M2´SO4 = Meq. of M2´CO3
∴ 71
W42 SO´M × 1000 =
10659.02× × 1000
or 42 SO´MW = 0.7904 g
∴ Total mass of sulphates = 42SOMW +
42 SO´MW = 0.6095 + 0.7904 = 1.3999 g
8. 0.9 g of a solid organic compound (molecular mass 90), containing carbon, hydrogen and oxygen, was heated with oxygen corresponding to a volume of 224 ml at STP. After combustion the total volume of the gases was 560 ml at STP. On treatment with potassium hydroxide, the volume decreased to 112 ml. Determine the molecular formula of the compound. [IIT-1972]
Sol. Given that, Mass of solid organic compound = 0.9 g Molecular mass of organic compound = 90 ∴ No. of moles of organic compound available
= 90
9.0 = 0.01
Volume of O2 taken = 224 ml Volume of O2 used = 224 – 112 = 112 ml 22400 ml O2 = 1 mol.
∴ 112 ml O2 = 22400112 = 0.005 mol
∴ At STP, no. of moles of O2 used = 22400112
= 0.005 mol Volume of CO2 obtained = 560 – 112 = 448 ml
∴ At STP, no. of moles of CO2 used = 22400
448
= 0.02 mol 0.01 mol organic compound yields = 0.02 mol CO2. ∴ 1 mol organic compound yields = 2 mol CO2 or 2 mol C ∴ The molecular formula of organic compound is
C2HyOz. The reaction is :
C2HyOz + 21 O2 → 2CO2 +
2y H2O
Equating no. of oxygen atoms,
z + 1 = 4 + 2y
or z = 3 + 2y
Molecular mass of C2HyOz = 2 × 12 + y × 1 + z × 16
Hence, 2 × 12 + y × 1 +
+
2y3 × 16 = 90
or y = 2
and z = 3 + 2y = 4
∴ The molecular formula of organic compound is C2H2O4.
9. (a) Write the intermediate steps for each of the
following reactions. (i) C6H5CHOHC ≡ CH →
+OH3 C6H5CH=CHCHO
(ii)
OH
H+
O CH3
(b) Show the steps to carry out the following transformations :
(i) Ethylbenzene → benzene (ii) Ethylbenzene → 2-phenylpropionic acid [IIT-1998] Sol. (a) (i)
C6H5CH(OH)C ≡ CH H+ C6H5CH – C ≡ CH
OH2+ –H2O
C6H5CH = C = CH C6H5CH – C ≡ CH ⊕ ⊕
OH–
C6H5CH = C = CHOH C6H5CH = CHCHO Tautomerism
XtraEdge for IIT-JEE 12 APRIL 2010
(ii)
OH H+ CH3
OH
CH3CH3 O O
H⊕
⊕
(b) (i)
C2H5
alk. KMnO4
Ethyl benzene
COOH
NaOH
Benzoic acid
–H2O
COONa
Sodium benzoate
NaOH
+CaO
Benzene
+ Na2CaO3
(ii)
CH2CH3
Br2/hv
CHBrCH3
Mg
Ether
CH3CH – MgBr
CO2
CH3CHCOOMgBrCH3CHCOOH
H2O/H+
Mg + Br
OH 2-Phenylpropionic acid
10. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998]
Sol. In H2S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4
Hence here sulphur is sp3 hybridised, so 16S = 1s2, 2s22p6,
44 344 21ionhybridisatsp
1z
1y
2x
2
3
p3p3p3s3
or
S
H HH H
S
Due to repulsion between lp - lp; the geometry of
H2S is distorted from tetrahedral to V-shape.
In PCl3, no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4
Hence, here P shows sp3-hybridisation 15P = 1s2, 2s22p6,
44 344 21ionhybridisatsp
1z
1y
1x
2
3
p3p3p3s3
P
or
P
Cl Cl
ClClCl
Cl
Thus due to repulsion between lp – bp, geometry is
distorted from tetrahedral to pyramidal.T
MATHEMATICS
11. 7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives ? [IIT-1985]
Sol. The possible cases are : Case I : A man invites 3 ladies and women invites 3
gentlemen ⇒ 4C3.4C3 = 16 Case II : A man invites (2 ladies, 1 gentleman) and
women invites (2 gentlemen, 1 lady) ⇒ (4C2.3C1).(3C1.4C2) = 324 Case III : A man invites (1 lady, 2 gentlemen) and
women invites (2 ladies, 1 gentleman) ⇒ (4C1.3C2).(3C2.4C1) = 144 Case IV : A man invites (3 gentlemen) and women
invites (3 ladies) ⇒ 3C3.3C3 = 1 ∴ Total number of ways = 16 + 324 + 144 + 1 = 485 12. Let n be a positive integer and (1 + x + x2)n = a0 + a1x + ..... + a2nx2n Show that a0
2 – a12 + ...... + a2n
2 = an. [IIT-1994] Sol. (1 + x + x2)n = a0 + a1x + .... + a2nx2n ...(1) Replacing x by –1/x. we obtain
n
2x1
x11
+− = a0 –
xa1 + 2
2
xa – 3
3
xa
+...+ n2n2
xa ..(2)
Now, a02 – a1
2 + a22 – a3
2 + ... + a2n2 = coefficient of
the term independent of x in [a0 + a1x + a2x2 + ... + a2nx2n]
+−+− n2
n2221
0 xa...
xa
xaa
= coefficient of the term independent of x in
(1 + x + x2)nn
2x1
x11
+−
But, R.H.S. = (1 + x + x2)nn
2x1
x11
+−
= n2
n2n2
x)1xx()xx1( +−++
XtraEdge for IIT-JEE 13 APRIL 2010
= n2
n222
x]x)1x[( −+
= n2
n242
x)xxx21( −++
= n2
n42
x)xx21( ++
Thus, a02 – a1
2 + a22 – a3
2 + ... a2n2
= coefficient of the term independent of x in
n2x1 (1 + x2 + x4)n
= coefficient of x2n in (1 + x2 + x4)n = coefficient of tn in (1 + t + t2)n = an 13. Solve for x the following equation : log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) [IIT-1987] Sol. log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) ⇒ log(2x + 3)(2x + 3) . (3x + 7) = 4 – log(3x + 7)(2x + 3)2 ⇒ 1 + log(2x + 3) )(3x + 7) = 4 – 2log(3x + 7) (2x + 3) Put log(2x + 3) (3x + 7) = y
⇒ y + y2 – 3 = 0 ⇒ y2 – 3y + 2 = 0
⇒ (y – 1) (y – 2) = 0 ⇒ y = 1 or y = 2 ⇒ log(2x + 3) (3x + 7) = 1 or log(2x + 3)(3x + 7) = 2 ⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2 ⇒ x = – 4 or 3x + 7 = 4x2 + 12x + 9 4x2 + 9x + 2 = 0 4x2 + 8x + x + 2 = 0 (4x + 1) (x + 2) = 0 x = – 2, –1/4 ∴ x = – 2, –4, –1/4 But, log exists only when, 6x2 + 23x + 21 > 0, 4x2 + 12x + 9 > 0, 2x + 3 > 0 and 3x + 7 > 0 ⇒ x > –3/2 ∴ x = –1/4 is the only solution. 14. Let f[(x + y)/2] = f(x) + f(y) / 2 for all real x and y,
if f´(0) exists and equals –1 and f(0) = 1, find f(2). [ΙΙΤ−1992]
Sol. f
+
2yx =
2)y(f)x(f + ∀ x, y ∈ R (given)
Putting y = 0, we get
f
2x =
2)0(f)x(f + =
21 [1 + f(x)] [Q f(0) = 1]
⇒ 2f(x/2) = f(x) + 1 ⇒ f(x) = 2f(x/2) – 1 ∀ x, y ∈ R ...(1) Since f´(0) = –1, we get
⇒ h
)0(f)h0(flim0h
−+→
= – 1 ⇒ h
1)h(flim0h
−→
Now, let x ∈ R then applying formula of differentiability.
f´(x) = h
)x(f)hx(flim0h
−+→
= h
)x(f2
h2x2flim
0h
−
+
→
= h
)x(f2
)h2(f)x2(f
lim0h
−+
→
= h
)x(f12h2f21
2x2f2
21
lim0h
−
−
+−
→
[using equation (1)]
= h
)x(f1)h(f21–)x(f221
lim0h
−−+
→
= h
1)h(flim0h
−→
= –1
Therefore f´(x) = – ∀ x ∈ R ⇒ ∫ )x´(f dx = ∫ −1 dx
⇒ f(x) = – x + k where k is a constant. But f(0) = 1, therefore f(0) = – 0 + k ⇒ 1 = k ⇒ f(x) = 1 – x ∀ x ∈ R ⇒ f(2) = – 1 15. If (a + bx)ey/x = x, then prove that
x32
2
dxyd =
2
ydxdyx
− [IIT-1983]
Sol. (a + bx).ey/x = x ...(1) Differentiating both sides, we get
(a + bx).ey/x.
−
2x
ydxdyx
+ bey/x = 1
⇒ x. 2x
ydxdyx
−
+ beb/x = 1, (using (1))
or dxdy –
xy + bey/x = 1,
Again differentiation both sides,
2
2
dxyd – 2x
ydxdyx −
+ bey/x.
−
2x
ydxdyx
= 0 ..(2)
from (2), 2
2
dxyd – 2x
ydxdyx −
−
xy
dxdy = 0
⇒ x32
2
dxyd =
2
ydxdyx
−
XtraEdge for IIT-JEE 14 APRIL 2010
1. Four infinite thin current carrying sheets are placed in YZ plane. The 2D view of the arrangement is as shown in fig. Direction of current has also been shown in the figure. The linear current density. i.e. current per unit width in the four sheets are I, 2I, 3I and 4I, respectively.
I Y
X
II III IV
a a aa The magnetic field as a function of x is best
represented by
(A)
B
X µ0I
a
2µ0I5µ0I
2a 3a 4a 5a
(B)
B
X µ0I
a
3µ0I
–µ0I 2a 3a 4a 5a
(C)
B
X +µ0I
a
4µ0I
–µ0I 2a 3a 4a 5a
(D)
B
X +µ0I
a
2µ0I
–µ0I2a 3a 4a 5a
2. Match the column
Column – I Column – II
(A) a charge particle is (P) Velocity of the moving in uniform particle may be electric and magnetic constant fields in gravity free space (B) a charge particle is (Q) Path of the particle moving in uniform may be straight line electric, magnetic and gravitational fields (C) a charge particle is (R) Path of the particle moving in uniform may be circular magnetic and gravitational fields (where electric field is zero) (D) A charge particle is (S) Path of the particle moving in only may be helical uniform electric field (T) None
3. Magnetic flux in a circular coil of resistance 10Ω changes with time as shown in fig. Cross indicates a direction perpendicular to paper inwards.
Match the following :
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma Director Academics, Jodhpur Branch
Physics Challenging Problems
Solut ions given in same issue
Set #12
XtraEdge for IIT-JEE 15 APRIL 2010
φ(Magnetic flux)
10
2 t(s)–10
68 10 14 16
× × × × × ×× × × × × ×× × × × × ×× × × × × × × × × × × × × × × × × ×
Column – I Column – II
(A) At 1s, induced current is (P) Clockwise (B) At 5s, induced current is (Q) Anticlockwise (C) At 9s, induced current is (R) Zero (D) At 15s, induced current is (S) 2A (T) None 4. A conducting rod of length l is moved at constant
velocity v0 on two parallel, conducting, smooth, fixed rails, which are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct ?
⊗v0 R ⊗B
(A) The thermal power dissipated in the resistor is
equal to the rate of work done by an external person pulling the rod
(B) If applied external force is doubled, then a part of the external power increases the velocity of the rod
(C) Lenz’s law is not satisfied if the rod is accelerated by an external force
(D) If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half
5. The x-z plane separates two media A & B of
refractive indices µ1 = 1.5 & µ2 = 2. A ray of light travels from A to B. Its directions in the two media
are given by unit vectors ,jbia1
∧∧→+=µ .jbic2
∧∧→+=µ
Then
(A) 34
ca
= (B) 43
ca
=
(C) 34
db
= (D) 43
db
=
6. Two converging lenses of the same focal length f are separated by distance 2f. The axis of the second lens is inclined at angle º60=θ with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Then
2f
60º
(A) Final image after all possible refraction will
formed at optical centre of first lens (B) Final image after all possible refraction will
formed at optical centre of second lens (C) Final image after all possible refraction will
formed at distance f from second lens
(D) Final image after all possible refraction will formed at distance f from first lens
7. If Cv for an ideal gas is given by Cv = (3 + 2T)R, where T is absolute temperature of gas, then the equation of adiabatic process for this gas is
(A) VT2 = constant
(B) VT3e-2T = C
(C) VT2e2T = constant
(D) VT3e2T = constant
8. The pressure of one mole of ideal gas varies according to the law P = P0 – αV2 where P0 & α are positive constant constants. The highest temperature that gas may attain -
(A) 2/1
00
3P
R3P2
α
(B) 2/1
00
3P
R2P3
α
(C) 2/1
00
3P
RP
α
(D) 2/1
00 PRP
α
XtraEdge for IIT-JEE 16 APRIL 2010
1.[C] The given circuit as an R-L-C series circuit when frequency of the source varies the impedance of the R-L-C series circuit varies and correspondingly the current in the circuit get varied
Impedance variation and current variation are shown in figure.
R = Z minimum
f1 f=fr f2 f
∆f = f2-f1
I
I0
f1 f=fr f2 f
∆f = f2-f1
2I
I 0=
At frequency f1 XC > XL Power factor – leading nature of circuit is capacitance
At frequency f2 XL > XC Power factor – leading nature of circuit is inducting
At frequency f1 and frequency f2 impedance Z= R.2
Because of the fact –
As )i........(RV
ZhmVI0 ==
ZV
2I0 ==±
⇒ ZV
22/V
=
⇒
2Rand
ZV
2RV
=
f1 < f < f2
2.[A] 12 fff −=∆ = B and width of R-L-C series circuit
L/R.
21π
=
3[C] At frequency f1 current
.Amp2102
2010200.
21
RV.
21
2I0 ====±
Watt less current
φ=µ sinII
.Amp10
21.210I ==µ
º45As =φ because 2
12R
RZ/Rcos ===φ
4[A] At frequency current .Amp2010200
RV
ZmVI ====
Potential difference across capacitor
VC = IC.XC = 1.XC = 20.XC
Charge on capacitor QC = C.VC = C. 20XC
= C. (2v) C
1ω
π
=π
=π
=ω
=51
)50(220
f22020
Coulomb)5( 1−π=
cb)5( 1−π=
5.[C] Longest wavelength
m59.0
60058.0350
fvv s
max =××
=+
=λ
6.[B] fvv
vfs
max −=
Hz607
58.0350350
=×−
=
7.[A] 345.5/346.0 × 600 = 599 Hz
8.[A] 2
2
22
2
dtyd
v1
dxyd&
dtdy
v1
dxdy
+=−=
Solution Physics Challenging Problems
Set # 11
8 Questions were Publ ished in March Issue
XtraEdge for IIT-JEE 17 APRIL 2010
XtraEdge for IIT-JEE 18 APRIL 2010
1.[C] Magnetic field due to infinite current carrying
sheet is given by ,2
JB 0µ
= where J is linear current
density.
2J0µ
II III
(a)
2J0µ
(b)
2J0µ
2J0µ
Fig. (a) and (b) represent the direction of magnetic field due to current carrying sheets. For x < a,
2)J4(
2)J3(
2)J2(J
2J
B 0000ttanresul
µ+
µ−
µ−
µ=
For a < x < 2a,
J2
)J4(2
)J3(2
)J2(2
JB 0
0000ttanresul µ−=
µ+
µ−
µ−
µ=
For 2a < x < 3a,
02
)J4(2
)J3(2
)J2(2
JB 0000
ttanresul =µ
−µ
−µ
+µ
=
So, the required curve is B
XO a 2a 3a 4a 5a
2. A → P,Q,S B → P,Q,R,S C → P,Q,R,S D → Q
(A) Velocity of the particle may be constant, if forces of electric and magnetic fields balance each other. Then, path of particle will be straight line. Also, path of particle may be helical if magnetic and electric fields are in same direction. But path of particle cannot be circular. Path can be circular if only magnetic field is present, or if some other forces is present which can cancel the effect of electric field.
(B) Here, all the possibilities are possible depending upon the combinations of the three fields.
(C) This situation is similar to part (i) (D) In a uniform electric field, path can be only
3. A → Q B → R C → P D → Q (A) At t = 1s, flux is increasing in the inward
direction, hence induced e.m.f. will be in anticlockwise direction.
(B) At t = 5s, there is no change in flux, so induced e.m.f. is zero
(C) At t = 9s, flux is increasing in upward direction hence induced e.m.f. will be in clockwise direction.
(D) At t = 15s, flux is decreasing in upward direction, so induced e.m.f. will be in anticlockwise direction.
4.[A, B, D] Rate of work done by external agent is
de/dt = BIL.dx/dt = BILv and thermal power dissipated in resistor = eI = (BvL) I clearly both are equal, hence (A). If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increase, hence (B). Since, I = e/R On doubling R, current and hence required power become half. Since, P = BILv Hence (D)
5.[A]
xz
n1sini = n2sinr
j
)j(2)j(5.1 21
∧→∧→×µ=×µ
]j)jdic[(2j)jbia(5.1∧∧∧∧∧∧
×+=×+
∧∧
= kc2ka5.1
34
5.120
ca
==
Solution Physics Challenging Problems
Set # 12
8 Questions Published in this Issue
XtraEdge for IIT-JEE 19 APRIL 2010
6.[A]
I2
+ –
f
f cos60º
I1
f sin60ºf
x u = -f cos60º f = +f
º60cosf
1v1
f1
−−=
f2
v1
f1
+=
v1
f2
f1
=−
v = -f
º60cosxf
=
xº60cos
f=
x = 2f ∴ final image will formed at optical centre of first lens.
7.[C] Cv = (3 + 2T)R dQ = dU + PdV adiabatic process dQ = 0 0 = Rn (3 + 2T)dT + PdV
dVV
nRTdT)T23(Rn0 ++=
∫ ∫
+
=− dTT
T23V
dV
– log V = 3 logT + 2T + C – logV – logT3 = 2T + C log VT3 = 2T + C VT3 = e2T VT3e-2T = C
8.[A] 20 VPP α−=
PV = RT
20 VP
VRT
α−=
RV
RVP
T3
0 α−=
0dVdT
=
0RV3
RP 2
0 =α
−
α
=3P
V 0 Now put V in T.
• Saturn’s rings are made up of particles of ice, dust and rock. Some particles are as small as grains of sand while others are much larger than skyscrapers.
• Jupiter is larger than 1,000 Earths.
• The Great Red Spot on Jupiter is a hurricane-like storm system that was first detected in the early 1600’s.
• Comet Hale-Bopp is putting out approximately 250 tons of gas and dust per second. This is about 50 times more than most comets produce.
• The Sun looks 1600 times fainter from Pluto than it does from the Earth.
• There is a supermassive black hole right in the middle of the Milky Way galaxy that is 4 million times the mass of the Sun.
• Halley’s Comet appears about every 76 years.
• The orbits of most asteroids lie partially between the orbits of Mars and Jupiter.
• Asteroids and comets are believed to be ancient remnants of the formation of our Solar System (More than 4 billion years ago!).
• Comets are bodies of ice, rock and organic compounds that can be several miles in diameter.
• The most dangerous asteroids, those capable of causing major regional or global disasters, usually impact the Earth only once every 100,000 years on average.
• Some large asteroids even have their own moon.
• Near-Earth asteriods have orbits that cross the Earth’s orbit. These could potentially impact the Earth.
• There are over 20 million observable meteors per day.
• Only one or two meteorites per day reach the surface of Earth.
• The largest found meteorite was found in Hoba, Namibia. It weighed 60 tons.
XtraEdge for IIT-JEE 20 APRIL 2010
1. A beam of length L, breadth b and thickness d when loaded by a weight Mg in the middle, a depression e is produced in it. By measuring this depression e, the value of Young's modulus of the material of beam can be calculated by using the expression
Y = edb4
LgM3
3
Following are the values of different physical quantities obtained in one set of observations on this experiment :
M = 1000 gms, L = 200 cm, b = 2.54 cm, d = 0.620 cm, e = 0.1764 cm. If M is measured by spring balance, L by metre scale,
b by vernier calipers, d by screw gauge and e by spherometer, then what will be the maximum possible percentage errors in Y ?
Sol. Given that Y = edb4
LgM3
3
Taking log on both sides of above equation, we get log Y = log M + log g + 3 log L – log 4 – log b – 3 log d – log e Differentiating above equations, we have :
YY∆ =
MM∆ + 3
LL∆ –
bb∆ – 3
dd∆ –
ee∆
In order to calculate maximum possible error, we shall convert negative sign into positive sign.
∴ YY∆ =
MM∆ + 3
LL∆ +
bb∆ + 3
dd∆ +
ee∆
Now, least counts of the different measuring instruments used in the experiment are as under :
Least count of spring balance = 5 gm i.e. ∆M = 5gm Least count of metre scale = 0.1 cm i.e. ∆L = 0.1 cm Least count of vernier callipers = 0.01 cm i.e. ∆b = 0.001 cm Least count of screw gauge = 0.001 cm i.e. ∆d = 0.001 cm Least count of spherometer = 0.005 cm i.e ∆e = 0.005 cm
∴ YY∆ =
10005 +
2001.03× +
54.201.0 +
62.0001.03× +
1764.0005.0
= 0.005 + 0.0015 + 0.00393 + 0.00484 + 0.02834 = 0.0436 or 4.36% Hence the maximum possible percentage error is
4.36%.
2. A small glass ball is released from rest from the top of a smooth incline plane of constant base b. find the angle of inclination of the plane for minimum time of motion of the glass ball.
A
BCb
θ
Sol. Let the angle of inclination be θ. If, the glass ball
reaches the bottom B of the inclined plane after a time, say t, the equation of motion along the plane is given as
AB = (VA)t + 21 (g sin θ)t2
A
B Cb
θ g
L = b sec θ
g sin θ
Since the glass ball is released from rest VA = 0,
hence AB = 21 (g sin θ)t2 ...(1)
Putting AB = (BC) sec θ = b sec θ, in equation (1), we obtain
t = θθcossing
b2 = θ2sing
b4
For t to be minimum, sin 2θ is maximum ∴ sin 2θ|max = 1
or, 2θ = 2π or, θ =
4π
3. Two particles, both of mass m, attract each other with
the force )r(Frr
= – rr 2α
where α is a positive constant. At a certain moment (t = 0), the distance between the particles is R, and their velocities are
−==
xv2vxvv
02
01
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' ForumPHYSICSS
XtraEdge for IIT-JEE 21 APRIL 2010
Assuming the two-particle system reaches a minimum of kinetic energy at a certain moment and at a certain finite distance between the particles (in the laboratory frame), find the distance between the particles at that moment and the value of that minimal kinetic energy.
Sol. For the sake of convenience, we will first solve the problem in the frame of the centre of mass. Then, we will transform the results into the laboratory frame. We will determine the velocity of the center of mass by :
cmvr
= xm2
mv2mv 00 − = – xv
21
0 ...(1)
This velocity remains constant since there are no external forces acting on the whole system. In the velocity of the centre of mass, the velocities of the particles are :
−=−=
=−=
xv23vvu
xv23vvu
0cm22
0cm11
rrr
rrr
...(2)
By definition of a centre of mass frame, the total momentum of the particles is zero. The kinetic energy in this system at t = 0 is :
K´ = 21 mu1
2 + 21 mu2
2 = 49 mv0
2 ...(3)
Therefore, the total energy in the center of mass frame at t = 0 is
E´ = K´ + U´ = 49 mv0
2 – Rα ...(4)
where we define R ≡ r(t = 0). Note that since the force is conservative, we have F
r = – ∇ u. The scalar
function is u = – α/r. The advantage of using the center of mass frame is
evident when one inspects the moment of arrival at a minimal distance, t0. At that moment, in this system, the two particles stop and reverse their directions. The kinetic energy, therefore, vanishes at t0 in the center of mass frame, or,
K´(t0) = 'minK = 0 ...(5)
Hence, E´(t0) = –minRα = E´ ...(6)
Plugging in the value of E´, we find :
Rmax = Rmv94
R420−α
α ...(7)
We now transform the centre of mass frame to the laboratory frame. Since Rmax is the relative distance between the two particles, it is unchanged by the transformation. Recall that distance is an invariant quantity of the transformations of displacement and / or rotation. Therefore,
Kmin – maxRα = E(0) = E =
21 mv0
2 + 21 m(2v0)2 –
Rα
....(8)
The kinetic energy in the laboratory frame is, therefore :
Kmin = 25 mv0
2 – Rα +
R4Rmv94 2
0
α−α
α
=
−
49
25 mv0
2 = 41 mv0
2 ...(9)
Another way of finding the minimal kinetic energy is by using the following formula :
K = K´ + 2cmMv
21 ...(10)
where K´ is the kinetic energy in the centre of mass frame, K is the energy in the laboratory frame, and M is the total mass of the system. In our case,
K = 0 + 2cmMv)m2(
21 = 2
0Mv41 ...(11)
Note : Generally, in transforming from system S to system S´ with relative velocity V
r, the kinetic
energy is transformed as :
K´ = K – M2
p2 +
2
MpV
2M
−
rr ..(12)
where M is the total mass and pr
is the total momentum in S. K´ is minimal in the center of mass
frame if we choose Vr
= cmvr
= Mpr
. We obtain :
K´ = K = M2
p2 ...(13)
4. A hot body is being cooled in air according to
Newton's law of cooling, the rate of fall of temperature being K times the difference of its temperature with respect to that of surroundings. Calculate the time after which the body will lose half the maximum heat it can lose. The time is to be counted from the instant t = 0.
Sol. According to Newton's law of cooling, we have
dtdθ = – K(θ – θ0)
where θ0 is the temperature of the surrounding and θ is the temperature of the body at time t. Suppose θ = θ1 at time t = 0.
Then, ∫θ
θ θ−θθ
1 0
d = –K ∫t
0dt or, log
01
0
θ−θθ−θ
= – Kt
or, θ – θ0 = (θ1 – θ0)e–Kt ...(1) The body continues to lose heat till its temperature
becomes equal to that of the surroundings. The loss of heat in this entire period is dQm = ms(θ1 – θ0).
This is the maximum heat the body can lose. If the body loses half this heat, the decreases in its temperature will be
ms2
dQm = 2
01 θ−θ
XtraEdge for IIT-JEE 22 APRIL 2010
If the body loses this heat in time t1, the temperature at t1 will be
θ1 – 2
01 θ−θ =
201 θ+θ
Putting these values of time and temperature in (1) :
2
01 θ+θ – θ0 = (θ1 – θ0) 1Kte−
or, 1Kte− = 21 or t1 =
K2log
5. In a certain region surrounding the origin of the
coordinates, →B = 5 × 10–4
→k T and
→E = k V/m. A
proton enters the fields at the origin with an initial
velocity →
0v = 2.5 × 105 i m/s. Describe the proton's motion and give its position after three complete revolutions.
Sol. The z-component of the force →F is a constant
electrical force. It produces a constant acceleration along z-axis given as
Z = 21 at2 =
21
meE t2
BE z
x
y The other component of the force F is a magnetic
force which provides the necessary centripetal force to keep the proton in a circular path of radius r(say). The period of revolution of the proton
T = 0vr2π As Fcp =
rmv2
0 = ev0B
∴ v0 = m
eBr Hence, T = eB
m2π
Since the particle (proton) moves in circular path having a period of revolution T in x, y plane and moves along z-axis with a constant acceleration a = eE/m, the path of the proton is a helix.
After three revolutions, putting t = 3T, we obtain
z =
meE
21 (3T)2 =
meET
29 2
Putting T = 2πm/eB, we get
z = 2
2
eBEm18π = 246
272
)105(106.11066.15)7/22(18
−−
−
××××××
= 37m
Interesting Science Facts
• The Universe contains over 100 billion galaxies. • Wounds infested with maggots heal quickly and
without spread of gangrene or other infection. • More germs are transferred shaking hands than
kissing. • The longest glacier in Antarctica, the Almbert
glacier, is 250 miles long and 40 miles wide. • The fastest speed a falling raindrop can hit you is
18mph. • A salmon-rich, low cholesterol diet means that
Inuits rarely suffer from heart disease. • Inbreeding causes 3 out of every 10 Dalmation
dogs to suffer from hearing disability. • The world’s smallest winged insect, the Tanzanian
parasitic wasp, is smaller than the eye of a housefly.
• If the Sun were the size of a beach ball then Jupiter would be the size of a golf ball and the Earth would be as small as a pea.
• It would take over an hour for a heavy object to sink 6.7 miles down to the deepest part of the ocean.
• There are more living organisms on the skin of each human than there are humans on the surface of the earth.
• The grey whale migrates 12,500 miles from the Artic to Mexico and back every year.
• Quasars emit more energy than 100 giant galaxies. • Quasars are the most distant objects in the
Universe. • The Saturn V rocket which carried man to the
Moon develops power equivalent to fifty 747 jumbo jets.
• Koalas sleep an average of 22 hours a day, two hours more than the sloth.
• Light would take .13 seconds to travel around the Earth.
• Neutron stars are so dense that a teaspoonful would weigh more than all the people on Earth.
• One in every 2000 babies is born with a tooth. • Every hour the Universe expands by a billion
miles in all directions. • Somewhere in the flicker of a badly tuned TV set
is the background radiation from the Big Bang. • The temperature in Antarctica plummets as low
as -35 degrees Celsius. • Space debris travels through space at over 18,000
mph.
XtraEdge for IIT-JEE 23 APRIL 2010
Calorimetry : The specific heat capacity of a material is the amount
of heat required to raise the temperature of 1 kg of it by 1 K. This leads to the relation
Q = ms θ where Q = heat supplied, m = mass, θ = rise in
temperature. The relative specific heat capacity of a material is the
ratio of its specific heat capacity to the specific heat capacity of water (4200 J kg–1K–1).
Heat capacity or thermal capacity of a body is the amount of heat required to raise its temperature by 1 K. [Unit : J K–1]
Thus heat capacity = Q/θ = ms
Also dtdθ =
ms1 ×
dtdQ
i.e., the rate of heating (or cooling) of a body depends inversely on its heat capacity.
The water equivalent of a body is that mass of water which has the same heat capacity as the body itself. [Unit : g or kg] This is given by
W = ws
sm×
where m = mass of body, s = specific heat capacity of the body, sw = specific heat capacity of water.
Principle of Calorimetry : The heat lost by one system = the heat gained by another system. Or, the net heat lost or gainsed by an isolated system is zero.
It system with masses m1, m2, ...., specific heat capacities s1, s2, ...., and initial temperatures θ1, θ2, .... are mixed and attain an equilibrium temperature θ then
θ = ´ms
msΣ
θΣ , for equal masses θ = ss
ΣθΣ
Newton's law of cooling : The rate of loss of heat from a body in an
environment of constant temperature is proportional to the difference between its temperature and that of the surroundings.
If θ = temperature of the surroundings then
– msdtdθ = C´(θ – θ0)
where C´ is a constant that depends on the nature and extent of the surface exposed. Simplifying
dtdθ = –C(θ – θ0) where C =
ms´C = constant
Kinetic theory of gases :
The pressure of an ideal gas is given by p = 31 µnC2
where µ = mass of each molecule, n = number of molecules per unit volume and C is the root square speed of molecules.
p = 31
ρC2 or pV = 31 mC2
where ρ is the density of the gas and m = mass of the gas.
Root Mean Square Speed of Molecules : This is defined as
C = N
C...CCC 2N
23
22
21 ++++
where N = total number of molecules. It can be obtained through these relations
C = ρp3 =
MRT3
Total Energy of an ideal gas (E) : This is equal to the sum of the kinetic energies of all
the molecules. It is assumed that the molecules do not have any potential energy. This follows from the assumption that these molecules do not exert any force on each other.
E = 21 mC2 =
23
Mm RT =
23 pV
Thus, the energy per unit mass of gas = 21 C2
The energy per unit volume = 23 p
The energy per mole = 23 pV =
23 RT
Calorimetry, K.T.G., Heat transfer
PHYSICS FUNDAMENTAL FOR IIT-JEE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
XtraEdge for IIT-JEE 24 APRIL 2010
Perfect gas equation : From the kinetic theory of gases the equation of an
ideal gas is pV = RT for a mole
and pV = Mm RT for any mass m
Avogadro number (N) and Boltzmann constant (k) : The number of entities in a mole of a substance is
called the Avogadro number. Its value is 6.023 × 1023 mol–1.
The value of the universal gas constant per molecular is called Boltzmann constant (k). Its value is 1.38 × 10–23 J K–1.
Degrees of Freedom : Principle of equipartition of energy : The number of ways in which energy may be stored
by a system is called its degrees of freedom. Principle of Equipartition of Energy : This
principle states that the total energy of a gas in thermal equilibrium is divided equally among its degrees of freedom and that the energy per degree of freedom is kT/2 where T is the temperature of the gas. For a monoatomic atom the number of degrees of freedom is 3, for a diatomic atom it is 5, for a polyatomic atom it is 6.
Hence the energy of a mole of a monoatomic gas is
µ = N
× kT
213 =
23 RT
Which is the same as that given by the kinetic theory. For a mole of diatomic gas µ
= N
× kT
215 =
25 RT
For a mole of polyatomic gas µ
= N
× kT
216 = 3RT
When the irrational degrees of freedom are also taken into account, the number of degrees of freedom
= 6n – 6 for non-linear molecules = 6n – 5 for linear molecules where n = number of atoms in a molecule. Kinetic Temperature : The kinetic temperature of a moving particle is the
temperature of an ideal gas in thermal equilibrium whose rms velocity equals the velocity of the given particle.
Maxwellian distribution of velocities : In a perfect gas all the molecules do not have the
same velocity, rather velocities are distributed among them. Maxwell enunciated a law of distribution of velocities among the molecules of a perfect gas. According to this law, the number of molecules with
velocities between c and c + dc per unit volume is
dn = 4πna3 2bce− c2 dc where
b = kT2m and a =
kT2mπ
and the number of molecules with the velocity c per unit volume is
nc = 4πna3 2bce− c2 The plot of nc and c is shown in the figure. The
velocity possessed by the maximum number of molecules is called the most probable velocity
α c Crms α = m/kT2 The mean velocity
c = πm/kT8 and vrms = πm/kT3
Conduction : The transfer of heat through solids occurs mainly by
conduction, in which each particle passes on thermal energy to the neighboring particle but does not move from its position. Very little conduction occurs in liquids and gases.
A
θ1 θ2
d Consider a slab of area A and thickness d, whose
opposite faces are at temperature θ1 and θ2 (θ1 > θ2). Let Q heat be conducted through the slab in time t.
Then Q = λA
θ−θd
21 t
where λ = thermal conductivity of the material. This has a fixed value for a particular material, being
large for good conductors (e.g., Cu, Ag) and low for insulators (e.g., glass, wood).
Heat Current : The quantity Q/t gives the heat flow per unit time, and is called the heat current.
In the steady state, the heat current must be the same across every cross-section. This is a very useful principle, and can be applied also to layers or slabs in contact.
tQ = – λA
dxdθ where the quantity
dxdθ =
d21 θ−θ is
called the temperature gradient.
XtraEdge for IIT-JEE 25 APRIL 2010
Unit of λ : Different units are used, e.g., cal cm s ºC–1, cal m–1 s–1 ºC–1, jm´1 s–1 ºC–1. Convection : It is a process by which heat is conveyed by the
actual movement of particles. Particles closest to the source receive heat by conduction through the wall of the vessel. They rise up-wards and are replaced by colder particles from the sides. Thus, a circulation of particles is set up – hot particles constitute the upward current and cold particles, the side and downward current.
The transfer of heat by convection occurs only in fluids, and is the main mode of heat transfer in them. Most fluids are very poor conductors.
Radiation : Thermal Radiation : Thermal radiations are
electromagnetic waves of long wavelengths. Black Body : Bodies which absorb the whole of the
incident radiation and emit radiations of all wavelengths are called black bodies.
It is difficult to realize a perfect black body in practice. However, a cavity whose interior walls are dull black does behave like a black body.
Absorption : Every surface absorbs a part or all of the radiation falling on it. The degree of absorption depends on the nature and colour of the surface. Dull, black surfaces are the best absorbers. Polished, white surfaces absorb the least. The coefficient of absorption for a surface is
aλ = incidentradiationabsorbedradiation
The suffix λ denotes the wavelength of the radiation being considered, Clearly, aλ = 1 for a black body, for all values of λ.
Emission : Each surface emits radiation (radiates) continuously. The emissive power (eλ) is defined as the radiation emitted normally per second per unit solid angle per unit area, in the wave-length range λ and λ + dλ. Clearly, the emissive power of a black body (denoted by Eλ) is the maximum.
Kirchhoff's Law : According to this law, for the same conditions of temperature and wavelength, the ratio eλ /aλ is the same for all surfaces and is equal to Eλ. This simply means that good absorbers are good emitters. Hence, a black body is the best emitter, and a polished white body, the poorest emitter.
Prevost's Theory of Exchanges : All bodies emit radiations irrespective of their temperatures. They emit radiations to their environments and receive radiations from their environments simultaneously. In the equilibrium state the exchange between a body and the environment of energy continues in equal amounts.
Stefan-Boltzmann Law : If a black body at an absolute temperature T be surrounded by another black body at an absolute temperature T0, the rate of loss of radiant energy per unit area is
E = σ(T4 – T04)
where σ is a constant called Stefan constant and its value is 5.6697 × 10–8 W m–2 K–4
The total energy radiated by a black body at an absolute temperature T is given by
E = σT4 × surface area × time Note : Remember that rate of generation of heat by
electricity is given by H = I2 R or R
V2or VI Js–1 or W.
1. An earthenware vessel loses 1 g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contains 9.5 kg of water. Find the time required for the water in the vessel to cool to 28ºC from 30ºC. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g–1.
Sol. Here water at the surface is evaporated at the cost of the water in the vessel losing heat.
Heat lost by the water in the vessel = (9.5 + 0.5) × 1000 × (30 – 20) = 105 cal Let t be the required time in seconds. Heat gained by the water at the surface = (t × 10–3) × 540 × 103
(Q L = 540 cal g–1 = 540 × 103 cal kg–1)
∴ 105 = 540t or t = 185 s = 3 min 5s
2. 15 gm of nitrogen is enclosed in a vessel at temperature T = 300 K. Find the amount of heat required to double the root mean square velocity of these molecules.
Sol. The kinetic energy of each molecule with mass m is given by
21 m 2
rmsv = 23 kT ...(1)
If we want to increase the r.m.s. speed to η times, then the temperature has to be raised to T´. Then,
2rmsmv
21 =
23 kT´ or
21 mη2 2
rmsv = 23 kT´ ...(2)
From eqs. (1) and (2), T´ = η2T ...(3)
Solved Examples
XtraEdge for IIT-JEE 26 APRIL 2010
The internal energy of n molecules at temperature T is given by
U = 25 nRT
Similarly, U´ = 25 nRT´
∴ Change in internal energy ∆U = 25 nR[T´ – T]
or ∆U = 25 nRT[η2 – 1]
= 25
Mm RT[η2 – 1]
= 25
2815 (8.31) (300) [4 – 1] = 104 J
3. 10 gm of oxygen at a pressure 3 × 105 N/m2 and
temperature 10ºC is heated at constant pressure and after heating it occupies a volume of 10 litres (a) find the amount of heat received by the gas and (b) the energy of thermal motion of gas molecules before heating.
Sol. (a) The states of the gas before and after heating are
PV1 = µM RT1 and PV2 =
µM RT2
Solving these equations for T2, we have
T2 = MR
PµV2 = )1031.8)(1010()103)(1010(32
33
53
×××××
−
−
= 1156 K
Now T2 – T1 = 1156 – 283 = 873 K The amount of heat received by the gas is given by
∆Q = µM CP(T2 – T1)
= 32
8731008.29)1010( 33 ××× −
= 7.9 × 103 J (b) The energy of the gas before heating
E1 = µM ×
2i × RT1
where i = number of degrees of freedom = 5 (for oxygen)
= 322
)283)(1031.8(5)1010( 33
×××× −−
= 1.8 × 103 J
4. A slab of stone of area 3600 sq cm and thickness 10 cm is exposed on the lower surface of steam 100ºC. A block of ice at 0ºC rests on upper surface of the slab. In one hour 4800 gm of ice is melted. Calculate the thermal conductivity of the stone.
Sol. The quantity of heat Q passing across the stone is given by
Q = d
t)TT(KA 21 −
Here A = 3600 sq. cm = 0.36 m2 d = 10 cm = 0.10 m, (T1 – T2) = 100 – 0 = 100ºC and
t = 1 hour = 3600 sec.
∴ Q = 10.0
360010036.0K ××× kilo-calories ...(1)
Now heat gained by the ice in one hour = mass of the ice × latent heat of ice = 4.8 × 80 kilo calories ...(2) From eqs. (1) and (2)
4.8 × 80 = 10.0
360010036.0K ×××
or K = 360010036.010.0808.4
××××
= 3 × 10–4 kilo cal m–1(ºC)–1s–1 5. A flat bottomed metal tank of water is dragged along
a horizontal floor at the rate of 20m/sec. The tank is of mass 20 kg and contains 1000 kg of water and all the heat produced in the dragging is conducted to the water through the bottom plate of the tank. If the bottom plate has an effective area of conduction 1 m2 and the thickness 5 cm and the temperature of water in the tank remains constant at 50ºC, calculate the temperature of the bottom surface of the tank, given the coefficient of friction between the tank and the floor is 0.343 and K for the material of the tank is 25 cal m–1 s–1 K–1.
Sol. Frictional force = µ m g = 0.343 × (1000 + 20) × 9.81 = 3432 N The rate of dragging, i.e., the distance travelled in
one second = 20 m. ∴ Work done per second = (3432 × 20) Nm/sec. This work done appears as heat at the bottom plate of
the tank. Hence
H = 18.4
203432× cal/sec
But H = d
)TT(KA 21 − (Q t = 1 sec)
Now 18.4
203432× = 05.0
)TT(125 21 −××
∴ T1 – T2 = 12518.405.0203432
×××× = 32.84
Temp. of bottom surface T1 = 50 + 32.84 = 82.84ºC
XtraEdge for IIT-JEE 27 APRIL 2010
Atomic Structure :
According to Neil Bohr's hypothesis is the angular momentum of an electron is quantised.
mvr = n
π2h or L = n
π2h
2πr = nλ
vn = Znmr2h
π =
137
c × nz ms–1
rn =
π 22
2
mke4h
Zn 2
= 0.529Z
n 2Å where k =
041πε
fn =
hrke2
× n1 =
n1058.6 15× Hz
K.E. = 21
rZke 2
; P.E. = rke2− × Z; T.E. = –
r2ke2
× Z
T.E. = 2
2
nZ6.13− ev/atom where –13.6
= Ionisation energy
⇒ +T.E. = 2
.E.P+ = – K.E.
Note : If dielectric medium is present then εr has to be taken into consideration.
cv =
λ1 = v =
−
ε 22
21
320
24
n1
n1
ch8zme
= RZ2
− 2
221 n
1n1 =
hp =
hmv
n = ∞ n = 7 n = 6 n = 5 n = 4
n = 3
n = 2
n = 1
Kδ
Kγ
Kβ
Lγ
Lβ
Lα
Balmer (Visible)
Limiting line of Lyman series
Lyman Series (U.V. rays)
Paschen (I.R.)
Brackett (I.R.)
Pfund (I.R.)
–0.85 eV
–1.5 eV
–3.4 eV
–13.6 eV
The maximum number of electrons that can be accommodated in an orbit is 2n2.
X-rays : When fast moving electron strikes a hard metal,
X-rays are produced. When the number of electrons striking the target metal increases, the intensity of X-rays increases. When the accelerating voltage/kinetic energy of electron increases λmin decreases. X-rays have the following properties :
(a) Radiations of short wavelength (0.01 Å – 10Å); high pentrating power; having a speed of 3 × 108 m/s in vacuum.
λmin λ
Inte
nsity
Continuous spectrum (Varies & depends on accelerating voltage)
Kβ
Kα
Lγ Lβ
Lα
Characteristic spectrum (fixed for a target material)
(b) λmin = eVhc =
E.Khc =
V12400 Å
(c) λ1 = R(Z – b)2
− 2n
11
b = 1 for k-line transfer of electron
(d) Moseley law ν = a(z – b)
R = R0A1/3 where R0 = 1.2 × 10–15 m R = radius of nucleus of mass number A. * Nucleus density is of the order of 1017 kg/m3
Isomers are nuclides which have identical atomic number and mass number but differ in their energy states.
Nucleon
energybindingNuclear = Amc2∆
where ∆m = mass defect
= A
c]Mm)ZA(Zm[ 2np −−+
Atomic Structure, X-Ray & Radio Activity
PHYSICS FUNDAMENTAL FOR IIT-JEE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
XtraEdge for IIT-JEE 28 APRIL 2010
* The binding energy per nucleon is small for small nuclei.
* For 2 < A < 20, there are well defined maxima which indicate that these nuclei are more stable.
* For 30 < A < 120 the average B.E./A is 8.5 MeV / nucleon with a peak value of 8.8 MeV for Iron.
* For A > 120, there is a gradual decreases in B.E./nucleon.
* More the B.E./A, more is the stability.
Radioactivity :
β particles are electrons emitted from the nucleus. (n → p + β)
(a) N = N0e–λt
(b) dtdN− = λN where
dtdN = activity level
(c) N = N0
n
21
= N0
2/1Tt
21
⇒ A = A0
n
21
where A = activity level
(d) T1/2 = λ693.0
(e) τ = λ1
(f) τ = 1.4 T1/2
(g) t = λ303.2 log10 N
N0 = λ303.2 log10 A
A0
= λ303.2 log
mm0
(h) If a radioactive element decays by simultaneous
emission of two particle then dtdN− = λ1N + λ2N
The following parameters remain conserved during a nuclear reaction
(a) linear momentum (b) Angular momentum (c) Number of nucleons (d) Charge (e) The energy released in a nuclear reaction X + P → Y + Z + Q
Q = [mx + mp) – (my + mz)]c2 = ∆m × c2 Q = ∆m × 931 MeV
(f) In a nuclear fusion reaction small nuclei fuse to give big nuclei whereas in a nuclear fusion reaction a big nuclei breaks down.
Thermal neutrons produce fission in fissile nuclei. Fast moving neutrons, when collide with atoms of comparable masses, transfer their kinetic energy to colliding particle and slow down.
According to Doppler's effect of light λλ∆ =
cv
Power, P = tE =
tnhν =
tnhcλ
η = putInputout
1. The energy of an excited hydrogen atom is –3.4 eV.
Calculate the angular momentum of the electron according to Bohr theory.
Sol. The energy of the electron in the nth orbit is
En = – 2n6.13 eV
Here, – 2n6.13 = –3.4
or n = 2
Angular momentum = π2
nh = 14.321063.62 34
××× −
= 2.11 × 10–34 Js.
2. The wavelength of the first member of the Balmer series in the hydrogen spectrum is 6563 Å. Calculate the wavelength of the first member of the Lyman series.
Sol. For the first member of the Balmer series
λ1 = R
− 22 3
121 =
36R5 ...(1)
For the first member of the Lyman series
´
1λ
= R
− 22 2
111 =
4R3 ...(2)
Dividing Eq. (1) by Eq. (2)
λλ´ =
33645
×× =
275
or λ´ = 275
λ = 275 × 6563 = 1215 Å
Solved Examples
XtraEdge for IIT-JEE 29 APRIL 2010
3. Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum ? Find the longest wavelength amongst these.
Sol. Energy the radiation quantum
E = hv = λhc = 1910
834
106.1106.970103106.6
−−
−
××××××
= 12.75 eV Energy of the excited sate En = – 13.6 + 12.75 = – 0.85 eV
Now, we know that En = – 2n6.13
or n2 = –nE6.13 =
85.06.13
−− = 16
or n = 4 The number of possible transition in going to the
ground state and hence the number of different wavelengths in the spectrum will be six as shown in the figure.
n 4 3
2
1 The longest wavelength corresponds to minimum
energy difference, i.e., for the transition 4 → 3.
Now E3 = – 236.13 = – 1.51 eV
max
hcλ
= E4 – E3
or λmax = 19
834
106.1)85.051.1(103106.6
−
−
××−×××
= 18.75 × 10–7m = 18750 Å
4. X-rays are produced in an X-ray tube by electrons
accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.
Sol. Initial kinetic energy of the electron = 50.0 keV Energy of the photon produced in the first collision,
E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon
λ1 = 1E
hc = 319
834
105.12106.1103106.6×××
×××−
−
= 0.99 × 10–10 m = 0.99 Å Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision , E3 = 12.5 – 0 = 12.5 keV This is same as E2. Therefore, wavelength of this
photon, λ3 = λ2 = 0.99 Å
5. In an experiment on two radioactive isotopes of an elements (which do not decay into each other), their mass ratio at a given instant was found to be 3. The rapidly decaying isotopes has larger mass and an activity of 1.0 µCi initially. The half lives of the two isotopes are known to be 12 hours and 16 hours. What would be the activity of each isotope and their mass ratio after two days ?
Sol. We have, after two days, i.e., 48 hours,
N1 = 4
01 2
1N
= 0
1N /16
N2 = 3
02 2
1N
= 0
2N /8
Mass ratio = 2
1
NN = 0
2
01
NN .
168 =
16283× =
23
Now, 01A = λ1
01N = 1.0 µCi
After two days,
A1 = λ1N1 = λ101N /16 = 0
1A /16 = (1/16)µCi
A2 = λ2N2 = λ202N /8
But 1
2
λλ =
2
1
TT =
1612 =
43
or λ2 = 43
λ1
A2 =
λ14
3 ×
0
1N31 ×
81
= 321
λ101N =
321 0
1A
= (1/32) µCi
XtraEdge for IIT-JEE 30 APRIL 2010
Halogenation of Benzene : Benzene does not react with bromine or chlorine
unless a Lewis acid is present in the mixture. (As a consequence, benzene does not decolorize a solution of bromine in carbon tetrachloride.) When Lewis acids are present, however, benzene reacts readily with bromine or chlorine, and the reactions give bromobenzene and chlorobenzene, respectively, in good yields :
+ Cl2
FeCl3
25ºC
Cl + HCl
Chlorobenzene (90%)
+ Br2FeBr3
heat
Br+ HBr
Bromobenzene (75%) The Lewis acids most commonly used to effect
chlorination and bromination reactions are FeCl3, FeBr3, and AlCl3, all in the anhydrous form.
A mechanism for the reaction : Electrophillic Aromatic Bromination : Step 1
Br – Br : + FeBr3 → :Br – Br – FeBr3
→ :Br+ + :Br – FeBr3
–
–
+
Bromine combines with FeBr3 to form a complex that dissociates
to form a positive bromine ion and FeBr4–
Step 2
+ Br:
slow
HBr:
HBr:
HBr:
Arenium ion The positive bromine ion attacks benzene to
form an arenium ion
+
+
+ +
Step 3
HBr: + H – Br: + FeBr3
Br:
A proton is removed from the arenium ionto become bromobenzene
+
–:Br – FeBr3
The function of the Lewis acid can be seen in step 1. The ferric bromide reacts with bromine to produce a positive bromine ion, Br+ (and FeBr4
–). In step 2 this Br+ ion attacks the benzene ring to produce an arenium ion. Then, finally in step 3 a proton is removed from the arenium ion by FeBr4
–. This results in the formation of bromobenzene and hydrogen bromide the products of the reaction. At the same time this step regenerates the catalyst, FeBr3.
Nucleophilic Aromatic Substitution through an Elimination – Addition Mechanism : Benzyne Although aryl halides such as chlorobenzene and
bromobenzene do not react with most nucleophiles under ordinary circumstances, they do react under highly forcing conditions. Chlorobenzene can be converted to phenol by heating it with aqueous sodium hydroxide in a pressurized reactor at 350ºC .
+ NaOH
Cl
350ºCH2O
ONa
PhenolH3O+
OH
Bromobenzene reacts with the very powerful base,
NH2– , in liquid ammonia :
+ K :NH2
Br
+ – -33ºCNH3
+ KBr
NH2
Aniline These reactions take place through an elimination –
addition mechanism that involves the formation of an interesting intermediate called benzyne (or dehydrobenzene). We can illustrate this mechanism with the reaction of bromobenzene and amide ion.
In the first step, the amide ion initiates an elimination by abstracting one of the ortho protons because they are the most acidic. The negative charge that develops on the ortho carbon is stabilized by the inductive effect of the bromine. The anion then loses a bromide ion. This elimination produces the highly unstable, and thus highly reactive, benzyne. Benzyne then reacts with any available nucleophile (in this case, an amide ion) by a two-step addition reaction to produce aniline.
Organic Chemistry
Fundamentals
AROMATIC HYDROCARBON
KEY CONCEPT
XtraEdge for IIT-JEE 31 APRIL 2010
The Benzyne Elimination – Addition Mechanism :
Br(–NH3)
Benzyne
(–Br–)
H
Br
(or dehydrobenzene)
:NH2
–
NH2
–
:NH3
NH2
H+ :NH2
:NH2 –
–
–
Addition
Elimination
Evidence for an elimination-addition mechanism : When 14C-labeled (C*) chlorobenzene is treated with
amide ion in liquid ammonia, the aniline that is produced has the label equally divided between the 1 and 2 positions.
Cl K+NH2
– * NH2–
NH3
NH2*
NH2
*(50%)
(50%)AdditionElimination
*
When the ortho derivative 1 is treated with sodium
amide, the only organic product obtained is m-(trifluoromethyl) aniline :
Cl
NaNH2
CF3
NH3 NH2
CF3
m-(Trifluoromethy)aniline1
This result can also be explained by an elimination –
addition mechanism. The first step produces the benzyne 2 :
Cl
NaNH2
CF3
NH3 + Cl–
CF3
1 2
This benzyne then adds an amide ion in the way that produces the more stable carbanion 3 rather than the less stable carbanion 4 :
CF3
:NH2
NH2
NH2
4
CF3
CF3
–
NH2 +:NH2
CF3
–
–
:NH3
2
3
Less stable carbanion
More stable carbanion (The negative charge is closer to the
electronegative trifluoromethyl group)
–
Carbanion 3 then accepts a proton from ammonia to
form m-(trifluoromethyl) aniline. Carbanion 3 is more stable than 4 because the carbon
atom bearing the negative charge is closer to the highly electronegative trifluoromethyl group. The trifluoromethyl group stabilizes the negative charge through its inductive effect. (Resonance effects are not important here because the sp2 orbital that contains the electron pair does not overlap with the π orbitals of the aromatic system.)
The Birch Reduction : Benzene can be reduced to 1, 4-cyclohexadiene by
treating it with an alkali metal (sodium, Lithium, or potassium) in a mixture of liquid ammonia and an alcohol.
Na
NH3, EtOHBenzene 1, 4-cyclohexadiene
A Mechanism for the Reaction : Brich Reduction : Na
etc.– –
Benzene Benzene radical anion
The first electron transfer produces a delocalized benzene radical anion.
EtOH
etc.
Cyclohexadienyl radical
H
H
H
H
Protonation produces a cyclohexadienyl radical(also a delocalized species)
Na
etc.
Cyclohexadienyl anion
H
HH
H
–
–
EtOHH
H
HH
1,4-CyclohexadieneTransfer of another electron leads to the formation of a delocalized
cyclohexadienyl anion, and protonation of this produces the 1,4-cyclohexadiene.
XtraEdge for IIT-JEE 32 APRIL 2010
Solubility : The amount of a solute, dissolved in a given volume
of a solvent (water) in 100 mL or in 1L to form a saturated solution at a given temperature is termed as the solubility of the solute.
Solubility Product : Salts like AgI, AgCl, PbI2, BaSO4, PbSO4 etc. are
ordinarily considered insoluble but they do possess some solubility. These are sparingly soluble salts. A saturated solution of sparingly soluble salt contains a very small amount of the dissolved salt. It is assumed that whole of the dissolved salt is present in the form of ions, i.e., it is completely dissociated. Consider a sparingly soluble salt like AgCl, the following equilibrium occurs between the undissolved solid salt and the silver and chloride ions in the saturated solution.
AgCl (s) Ag+ (aq) + Cl–(aq) Applying the law of mass action to the ionic
equilibrium,
K = )]s(AgCl[
]Cl][Ag[ −+
or K[AgCl(s) = [Ag+] [Cl–] The concentration of solid AgCl in the solid state i.e.
[AgCl(s)] is constant at a particular temperature, no matter how much solid is present in contact with the solution. It follows that
[AgCl(s)] = K´ = constant Hence, [Ag+] [Cl–] = KK´ = Ksp (constant) Ksp is termed as the solubility product. It is defined as
the product of the concentration of ions in a saturated solution of a salt at a given temperature. Consider, in general, the salt of the type AxBy which is dissociated as :
AxBy x Ay+ + y Bx– Applying law of mass action,
]BA[
]B[]A[
yx
yxxy −+
= K
when the solution is saturated, [Ax By] = K´ (constant) or [Ay+]x [Bx–]y = K[AxBy] = KK´ = Ksp (constant) Thus, solubility product is defined as the product of
concentrations of the ions raised to a power equal to the number of times the ions occur in the equation
representing the dissociation of the salt at a given temperature when the solution is saturated.
Solubility product is not the ionic product under all conditions but only when the solution is saturated. Ionic product has a broad meaning since it is applicable to all types of solutions, may be saturated or unsaturated.
Relationship between solubility and solubility product : The equilibrium for a saturated solution of any
sparingly soluble salt be expressed as : Ax By x Ay+ + y Bx– Thus, solubility product, Ksp = [Ay+]x [Bx–]y . Let 's' mole per litre be the solubility of the salt, then Ax By x Ay+ + y Bx– xs ys So KSP = [xs]x [ys]y = xx.yy(s)x+y Since the solubility of a salt varies with temperature,
the numerical value of Ksp for a salt changes with temperature; values usually recorded at 25ºC.
Common Ion Effect : The suppression of the degree of dissociation of a
weak acid or a weak base by the addition of a strong electrolyte containing a common ion. The common ion effect play an important role in the qualitative analysis.
Application of solubility product in qualitative analysis: Precipitation of sulphides of group II. Sulphides of
group II are precipitated by passing H2S gas through the solution of these cations in presence of dil HCl. H2S being a weak electrolyte ionizes only sligthtly, while HCl being a strong electrolyte is almost completely ionized.
H2S 2H+ + S2– ; HCl → H+ + Cl– Thus, the common ion effect takes place. As a result,
the degree of dissociation of H2S decreases sufficiently and the concentration of S2– ions in the solution becomes very small. But with this low concentration of second group and the sulphide ions exceeds the very low solubility products of their corresponding sulphides. Therefore, the cations of group II get precipitated as their insoluble sulphides.
On the other hand, the sulphides of the cations of the other groups (III, IV, V and Mg) are not precipitated under these conditions because their solubility products are quite high.
Physical Chemistry
Fundamentals
SOLUBILITY PRODUCT
KEY CONCEPT
XtraEdge for IIT-JEE 33 APRIL 2010
Precipitation of the hydroxides of group III : Hydroxides of group III are precipitated by adding an excess of solid NH4Cl to the solutions of these cations followed by the addition of excess of NH4OH. Being a weak electrolyte, NH4OH is only slightly ionised, whereas NH4Cl, being a strong electrolyte, ionizes almost completely to give at large concentration of NH4
+ ions. NH4OH NH4
+ + OH–; NH4Cl → NH4+ + Cl–
Due to the common ion effect, the degree of dissociation of NH4OH gets suppressed and hence the concentration of OH– ions in solution decreases appreciably. But even with this low conc. of OH– ions, the ionic products of the cations of group III and OH–ions exceed the low values of the solubility products of their corresponding metal hydroxides. As a result, the cations of group III get precipitated as their insoluble hydroxides.
On the other hand, cations of groups IV, V and Mg, which require a large conc. of OH– ions due to their high solubility products will not be precipitated.
Precipitation of sulphides of group IV. The sulphides of group IV are precipitated by passing H2S through ammoniacal solution of these cations.
Both H2S and NH4OH, being weak electrolytes, ionize only slightly as :
H2S 2H+ + S2–
NH4OH +4NH + OH–
The H+ ions and OH– ions combine to produce practically unionised molecules of water
H+ + OH– → H2O As a result, the above dissociation equilibrium
reactions get shifted in the forward direction, so that the concentration of S2– ions goes on increasing. Ultimately, the ionic product of the cations of group IV and S2– ions exceed the solubility products of their corresponding metal sulphides and hence get precipitated.
Precipitation of carbonates of group V : The carbonates of group V are precipitated by adding (NH4)2CO3 solution to the solution of these cations in the presence of NH4Cl and NH4OH. (NH4)2CO3, being a weak electrolyte ionises only slightly to give a small concentration of CO3
2– ions. (NH4)2CO3 2NH4
+ + CO32–
On the other hand, NH4Cl being a strong electrolyte, ionises almost completely to give a large concentration of +
4NH ions. Due to the common ion effect, the dissociation of (NH4)2CO3 is suppressed and hence the concentration of CO3
2– ions in the solution decreases considerably. But even with this low concentration of CO3
2– ions, the ionic products of these cations and CO3
2– ions exceed the low values of the solubility products of their corresponding metal carbonates and thus get precipitated.
However, under these conditions, Mg salts do not get precipitated as MgCO3 since its solubility product is comparatively high and thus requires a high concentration of CO3
2– ions for precipitation. The carbonates of Na+, K+ and +
4NH ions are also not precipitated because they are quite soluble.
The necessity of adding NH4OH arises due to the fact that (NH4)2CO3 solution usually contains a large amount of NH4HCO3. Thus, the cations of group V will form not only insoluble carbonates but soluble bicarbonates as well. As a result, the precipitation will not be complete. In order to convert NH4HCO3 to (NH4)2CO3, NH4OH is always added.
NH4HCO3 + NH4OH → (NH4)2CO3 + H2O Preferential precipitation of Salts : A solution contains more than one ion capable of
forming a precipitate with another ion which is added to the solution. For example, in a solution containing Cl–, Br–, and I– ions, if Ag+ ions are added, then out of the three, the least soluble silver salt is precipitated first. If the addition of Ag+ ions is continued, eventually a stage is reached when the next lesser soluble salt starts precipitating along with the least soluble salt and so on. If the stoichiometry of the precipitated salts is the same, then the salt with the minimum solubility product (and hence also the minimum solubility) will precipitate first followed by the salt of next higher solubility product and so on.
If the stoichiometry of the precipitated salts is not the same, then, from the solubility product data alone, we cannot predict which ion will precipitate first. Take, for example, a solution containing Cl– and CrO4
2–. Both these ions form precipitates with Ag+. Through the solubility product product of AgCl is larger than that of Ag2CrO4, yet it is AgCl (lesser soluble) which precipitates first when Ag+ ions are added to the solution. In order to predict which ion (Cl– or CrO4
2–) precipitates first, we have to calculate the concentration of Ag+ ions needed to start the precipitation through the solubility product data and the given concentration of Cl– or CrO4
2–. Since square root is involved in the expression for computing Ag+ for silver chromate, the quantity of Ag+ needed to start the precipitation of CrO4
2– is larger than that for Cl–. Hence, as AgNO3 is added to the solution, the minimum of the two concentrations of Ag+ to start the precipitation will be reached first and thus the corresponding ion (Cl– in this case) will be precipitated in preference to the other. During the course of precipitating, concentration of Cl– decreases and the corresponding concentration of Ag+ to start the precipitation increases. Its concentration eventually becomes equal to the value required for CrO4
2–. At this stage, practically the whole of Cl– ions have been precipitated. The addition of more of AgNO3 causes the precipitation of both the ions together.
XtraEdge for IIT-JEE 34 APRIL 2010
1. An organic compound (A), C4H9Cl, on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourise bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning.
Sol.
C4H9Cl
(A) (Alkyl halide)
C4H8 (C)
(Alkene)
Alc. KOH
∆;– HCl
C4H9OH (B)
(Alcohol)
Aq. KOH
∆; –KCl
Cu
∆; –H2O
We know that p-alcohol on heating with Cu gives
aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a t-alcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride.
Thus, (A) is
3
33
CH|
CHCCH|Cl
−− and (B), is
3
33
CH|
CHCCH|OH
−−
The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α - H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++
CH3 – C ≡ CH + H2O
Hg++
H+ CH3 – C = CH2
OH
CH3 – C – CH3
O
(D)
Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene.
CH3 – C = CH2
CH3
(C) (D) reacts with hydroxyl amine (NH2OH) to form
oxime (F).
C = O + H2 NOH
CH3
(D)CH3
–H2O C = NOHCH3
CH3(F)
Thus, (B) is
3
33
CH|
CHCCH|OH
−− and (A) is
3
33
CH|
CHCCH|Cl
−−
Reactions :
CH3 – C – CH3
Cl
CH3
Aq. KOH ∆; –KCl
CH3 – C – CH3
OH
CH3
Cu/
300º
C
–H2O
CH3 – C = CH2 + H2O
CH3
(A) (B)
(C)
Alc. KOH/∆ –KCl; –H2O
CH3 – C = CH2
CH3
(C)
CH3 – C = CH2(I) O3
(II) H2O/Zn C = O + H –C – H
CH3
CH3
O
(E)(D)
(C)CH3
∆
–H2OC = O + H2NOH
CH3
CH3 (F)(D)C = NOH
CH3
CH3
2HCHO + NaOH → CH3OH + HCOONa (E) (G) (H)
CH3 – C ≡ CH + H2OHg++
CH3 – C – CH3
O
(D)H+
UNDERSTANDINGOrganic Chemistry
XtraEdge for IIT-JEE 35 APRIL 2010
2. An organic compound (A) C7H15Cl on treatment with alcoholic KOH gives a hydrocarbon (B) C7H14. (B) on treatment with O3 and subsequent hydrolysis gives acetone and butyraldehyde. What are (A) and (B) ? Explain the reactions.
Sol. In numerical, following data are given :
halide Alkyl)A(
157 ClHC∆
→ KOH .Alc
Alkene)B(147HC → 3O
Ozonide3147 OHC
∆
→ Zn/OH2 CH3COCH3 + CH3CH2CH2CHO
The alkene contains seven carbon atoms. The position of C = C double bond can be located as follows :
C = O + O = C.CH2CH2CH3 CH3
CH3
H
–2[O]
CH3 – C = CH.CH2CH2CH3
CH3 Thus, alkene (B) is 2-methyl hexene-2 The ozonolysis reaction is as follows :
CH3 – C = CH.CH2CH2CH3
CH3 (B)
O3
CH3 – C CH – CH2CH2CH3
O O
CH3
O
H2O/Zn CH3 – C = O + O = C – CH2CH2CH3
CH3 H
∆
Since alkene (B) is produced by the removal of one
mol of HCl from alkyl halide (A) and thus (A) can be either (I) or (II).
CH3 – C – CH2CH2CH2CH3
Cl
CH3
(I) or
CH3 – CH – CH – CH2CH2CH3
CH3
(II)
Cl The dehydro halogenation reaction by(I) or (II) yeilds
CH3 – C – CH2CH2CH2CH3
Cl
CH3
KOH alc.
(I). CH3 – C = CH.CH2CH2CH3
CH3 Main product (Saytzeff's rule)
CH3 – CH – CH – CH2CH2CH3
CH3
KOH Alc.
(II) CH3 – C = CH.CH2CH2CH3
CH3 Main product (Saytzeff's rule)
Cl
Thus, both (I) and (II) give main product as 2-methyl hexene-2, hence (A) is either (I) or (II).
CH3 – C – CH2CH2CH2CH3
Cl
CH3
2-chloro-2-methyl hexane
(A)
CH3 – CH – CH – CH2CH2CH3
ClCH3
3-chloro-2-methyl hexane
or
CH3 – C = CH.CH2CH2CH3
CH32-methyl hexene-2
(B)
3. An unsaturated hydrocarbon (A), C6H10 readily gives
(B) on treatment with NaNH2 in liquid NH3. When (B) is allowed to react with 1-chloro propane, a compound (C) is obtained. On partial hydrogenation in the presence of Lindlar catalyst (C) gives (D), C9H18. On ozonolysis (D) gives 2, 2-dimethyl propanal and butanal. Give structures of (A), (B), (C) and (D) with proper reasoning.
Sol. The structure of compound (D) can be obtained by joining the products of ozonolysis.
CH3 – C – CH = O + O = CH.CH2CH2CH3
CH3
CH3
2,2-dimethyl propanal
–2[O]
Butanal
CH3 – C – CH = CH.CH2CH2CH3
CH3
CH3 2,2-dimethyl heptene-3 (D)
Ozonolysis equation of (D) is :
CH3 – C – CH = CHCH2CH2CH3
CH3
CH3
(D)
(I) O3
CH3 – C – CHO + CH3CH2CH2CHO
CH3
CH3
(II) H2O/Zn
Alkene (D) is obtained by the partial hydrogenation
of (C), thus (C) contains a – C≡C – triple bond at C3.
CH3 – C – C ≡ C – CH2CH2CH3
CH3
CH3
(C)
H2
CH3 – C – CH = CHCH2CH2CH3
CH3
CH3
Lindlar catalyst
(D)
XtraEdge for IIT-JEE 36 APRIL 2010
The starting compound (A) reacts with NaNH2 in presence of liquid NH3. It means it contains one –C≡CH at the terminal carbon, and, therefore gives a mono sodium derivative.
)A(106HC
3
2NH
NaNH → C4H9 – )B(
C ≡ C.Na
Compound (B) reacts with 1-chloro propane to give compound (C) as follows :
C4H9 – C ≡ C – Na + Cl – CH2CH2CH3 ∆
C4H9 – C≡C – CH2CH2CH3
–NaCl
(C)
(B) 1-chloro propane
But, (C) is CH3 – C – C ≡ C – CH2CH2CH3
CH3
CH3
Now, putting the value of C4H9 as a t-butyl radical, we have :
CH3 – C – C≡C – H
CH3
CH3 (A)
NaNH2 CH3 – C – C≡CNa + NH3
CH3
CH3 (B)
Hence,
CH3 – C – C≡CH
CH3
CH3 (3,3-dimethyl butyne-1)
(A)
CH3 – C – C≡CNa
CH3
CH3
(B)
CH3 – C – C≡C – CH2CH2CH3
CH3
CH3
(C)
CH3 – C – CH = CHCH2CH2CH3
CH3
CH3
(D)
4. A hydrocarbon (A) [C = 90.56%, V.D. = 53] was
subjected to vigrous oxidation to give a dibasic acid (B). 0.10 g of (B) required 24.10 ml of 0.05 N NaOH for complete neutralization. When (B) was heated strongly with soda-lime it gave benzene. Identify (A) and (B) with proper reasoning and also give their structures.
Sol. Determination of empirical formula of (A) :
Element % Atomic wt.
Relative no. of atoms Simplest ratio
C 90.56 12 12
56.90 = 7.55 55.755.7 = 1 or 4
H 9.44 1 144.9 = 9.44 55.7
44.9 = 1.25
or 5
The empirical formula of (A) = C4H5 Empirical formula weight = 48 + 5 = 53 Molecular weight = V.D. × 2 = 53 × 2 = 106
Hence, n = wt.Empirical wt.Molecular =
53106 = 2
Molecular formula = 2 × C4H5 = C8H10 The given equation may be outlined as follows :
C8H10
(A)
Vigrous oxidation6[O] C6H4
(B)
COOH
COOH+ 2H2O
Meq. of dicarboxylic acid = Meq. of NaOH
E
000,11.0 × = 24.1 × 0.05
Equivalent of acid = 83 Molecular wt. = Basicity × Equivalent weight = 2 × 83 = 166 Since (B) on heating with soda-lime gives benzene,
the C6H4 represents to benzene nucleus having two side chains, thus (B) is a benzene dicarboxylic acid.
There are three benzene dicarboxylic acids.
COOHCOOH
Phthalic acid
COOH
COOH Isophthalic acid
COOH
COOHTerphthalic acid
All the above three acids are obtained by the oxidation of respectively xylenes.
CH3
CH3
o-xylene
COOH COOH
6[O]+ 2H2O
CH3
CH3m-xylene
COOH
COOH
6[O]+ 2H2O
CH3
CH3p-xylene
COOH
COOH
6[O]+ 2H2O
XtraEdge for IIT-JEE 37 APRIL 2010
All the above three acids on heating with soda-lime yields only benzene.
COOH COOH
COOH COOH , ,
COOH
COOH
NaOH + CaO∆ + 2CO2
Of the three acids, one which on heating gives an anhydride, is o-isomer.
COOH CO CO COOH
O ∆
–H2O
One acid which on nitration gives a mono nitro compound is p-dicarboxylic acid.
COOH
COOH
NO2 HNO3
∆; H2SO4
COOH
COOH
One acid which on nitration gives three mono nitro compounds will be the m-isomer.
COOH
COOH NO2
HNO3
H2SO4
COOH
COOH
NO2
COOH
COOH
NO2
COOH
COOH
5. Two moles of an anhydrous ester (A) are condensed in presence of sodium ethoxide to give a β-keto ester (B) and ethanol. On heating in an acidic solution compound (B) gives ethanol and a β-keto acid (C). (C) on decarboxylation gives (D) of molecular formula C3H6O. Compound (D) reacts with sodamide to give a sodium salt (E), which on heating with CH3I gives (F), C4H8O, which reacts with phenyl hydrazine but not with Fehling reagent. (F) on heating with I2 and NaOH gives yellow precipitate of CHI3 and sodium propionate. Compound (D) also gives iodoform, but sodium salt of acetic acid. The sodium salt of acetic acid on acidification gives acetic acid which on heating with C2H5OH in presence of conc. H2SO4 gives the original ester (A). What are (A) to (F) ?
Sol. (i) Acetic acid on heating with C2H5OH gives original compound (A).
CH3COOH + C2H5OH ∆ → 42SOH
)A(523 HCOOCCH
+ H2O
(ii) CH3COOC2H5 (A) on heating with C2H5ONa undergoes Claisen condensation to give (B), which is aceto acetic ester.
CH3CO OC2H5 + H CH2COOC2H5 C2H5ONaReflux
(A)+ C2H5OH + CH3COCH2COOC2H5
(B) (iii) (B) on heating in acidic solution gives (C) and
ethyl alcohol.
)B(
5223 HCOOCCOCHCH + HOH →+H
)C(
23 COOHCOCHCH + C2H5OH
(iv) (C) on decarboxylation gives acetone (D).
)C(
23 COOHCOCHCH 2CO−
∆→ )D(
33COCHCH
(v) (D) reacts with NaNH2 to form sodium salt (E), which on heating with CH3I gives butanone (F).
)D(
33COCHCH + NaNH2 3NH−
∆→ )E(
23 NaCOCHCH
NaI–ICH3 →
)F(323 CHCOCHCH
(vi) )F(
323 CHCOCHCH + 3I2 + 4NaOH →∆
CHI3 + CH3CH2COONa + 3NaI + 3H2O
(vii) )D(
33COCHCH + 3I2 + 4NaOH →∆
CHI3 + CH3COONa + 3NaI + 3H2O
CH3COONa →HCl CH3COOH + NaCl
Thus, (A) CH3COOC2H5
(B) CH3COCH2COOC2H5
(C) CH3COCH2COOH
(D) CH3COCH3
(E) CH3COCH2Na
(F) CH3COCH2CH3
XtraEdge for IIT-JEE 38 APRIL 2010
1. Prove that, if n is a positive integer,
∫ −a
0
nx dxxe =
n !
++++− −
!na...
!2aa1e1
n2a
Also, deduce the value of ∫∞ −
0
nx dxxe
2. Let A ≡ (6, 5), B ≡ (2, –3) and C ≡ (–2, 1) be the
vertices of a triangle. Find the point P in the interior of the triangle such that ∆PBC is an equilateral triangle.
3. If Sn = nC1 + 2.nC2 + 3.nC3 + ....... + n nCn then find
∑=
n
1nnS . Also prove that
nC1 . (nC2)2 . (nC3)3.... (nCn)n ≤ 21n Cn
1n2
+
+.
4. Let z1, z2, z3 be three distinct complex numbers
satisfying |z1 – 1| = |z2 – 1| = |z3 – 1|; Let A, B and C be the
points represented in the argand plane corresponding to z1, z2 and z3 respectively. Prove that z1 + z2 + z3 = 3 if and only if ∆ABC is an equilateral triangle.
5. Let A ≡ (r, 0) be a point on the circle x2 + y2 = r2 and
D be a given point inside the circle. If BC be any arbitrary chord of the circle thorugh point D. Prove that the locus of the centroid of triangle ABC is a circle whose radius is less than r/3.
6. A number is chosen at random from the set 1, 2, 3, …......, 2006. What is the probability that it has no prime factor in common with 10 ! ?
7. Two vertices of a triangle are a – j3i + and j5i2 +
and its orthocenter is at ( j2i + ). Find the position
vector of third vertex. 8. Show that an equilateral triangle is a triangle of
maximum area for a given perimeter and a triangle of minimum perimeter for a given area.
9. The bottom of a tank with a capacity of 300 litres is
covered with a mixture of salt and some insoluble substance. Assuming that the rate at which the salt dissolves is proportional to the difference between the concentration at the given time and the concentration of a saturated solution (1 kg of salt per 3 litres of water) and that the given quantity of pure water disolves 1/3 kg of salt in 1 minute. Find the quantity of salt in solution at the expiration of one hour.
10. An isosceles triangle with its base parallel to the
major axis of the ellipse 9
x2+
3y2
= 1 is
circumscribed with all the three sides touching the ellipse. The least possible area of the triangle is.
`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions published in this issue
12Set
XtraEdge for IIT-JEE 39 APRIL 2010
1. Let z = x + iy so given 1 ≤ x or 1 ≤ r cos θ , if z = ∑eiθ Now,
iyx1iyx1
++−− = 22 y)x1(
)iyx1)(iyx1(++
−+−−
Real part 22
222
y)x1(y)x1(
++−− ≤ 0 as x ≥ 1 given and
imaginary part
22 y)x1(y)x1(y)x1(
+++−−− = 22 y)x1(
y2++
− ≤ 0
as y ≥ 0 given
so z1z1
+− ∩ 0 is true.
2. As ∠ POQ = 90º P
C
Q
O θ 90–θ
so CP = (OC) . tan (90 – θ) = (OC) cot θ & CQ = (OC) tan θ so CP. CQ = (OC)2 = r2 when r is the radius of circle.
3. f(0) = c f(1) = a + b+ c & f(–1) = a – b + c
solving these, a = 21 [f(1) + f(–1) – 2f(0)]
b = 21 [f(1) – f(–1)] & c = f(0)
so f(x) = 2
)1x(x + f(1) + (1 – x2) f(0) + 2
)1x(x − f(–1)
2 | f(x) | ≤ |x| |x + 1| + 2 |1 – x2| + |x| |x – 1|; as |f(1)|, |f(0)|, |f(–1)| ≤ 1. 2 |f(x)| ≤ |x| (x + 1) + 2(1 – x2) + |x| (1 – x) as x ∈ [–1, 1]
so 2 |f(x)| ≤ 2(|x| + 1 – x2) ≤ 2 . 45
so |f(x)| ≤ 45
Now, as g(x) = x2 f
x1
= 21 (1 + x) f(1) + (x2 – 1) f(0) +
21 (1 – x) f(–1)
so 2 | g(x) | ≤ |x + 1| + 2 |1 – x2| + |1 – x| 2 | g(x) | ≤ x + 1 + 2(1 – x2) + 1 – x as x ∈ [–1, 1] 2 |g(x)| ≤ –2x2 + 4 ≤ 4 |g(x)| ≤ 2
4.
−− ´y
ya2
y21a1
22
= 22 yaa
1
−+ .
22 ya2
´y.y2
−
− – y1 y´
a1 = y´
−
−−+−
− y1
ya)yaa(
y
yaa
y222222
= 2222
22222222
ya)yaa(ay
ya)yaa(aay)yaa(y´y
−−+
−−+−−−+
y(a + 22 ya − ) 22 ya −
= y´[y2a + y2 22 ya − – y2a – a2 22 ya − – a(a2 – y2)]
y (a + 22 ya − ) 22 ya − = –y´(a2 – y2) (a + 22 ya − )
y´ = – 22 ya
y
−
y´M = 22a β−
β−
y = xa 22 β−
β− ...(1)
y = β ...(2) locus of intersection of these two lines is
y = 22 ya
y
−
− x
a2 – y2 = x2 x2 + y2 = a2
MATHEMATICAL CHALLENGES SOLUTION FOR MARCH ISSUE (SET # 11)
XtraEdge for IIT-JEE 40 APRIL 2010
5. ∑= ++
−n
0r
rn
r
)1r()2r(2.C)2(
= )2n)(1n(2
1++ ∑
=
−n
0r
r)2( n+2Cr+2 (–2)2
= )2n)(1n(2
1++ ∑
=
+−n
0r
2r)2( n+2Cr+2
= )2n)(1n(2
1++
[(1 – 2)n + 2 –1 + 2(n + 2)]
= )2n)(1n(2
1++
[(–1) n + 2 + 2n + 3]
= 2n
1+
if n is odd
= 2n
1+
if n is even
6. In = 1
0
11n
1nxtanx
+
−+
– ∫ ++
+1
0 2
1n
)x1)(1n(x dx ...(1)
= )1n(4 +
π –
+−+
+ ∫−
dxx1
)11x(x1n
1 1
0 2
21n
= )1n(4 +
π – ∫ −
+
1
0
1n dxx1n
1 + ∫ ++
−1
0 2
1ndx
x1x
1n1
= )1n(4 +
π – 1
0
n
nx
1n1
+ + ∫ ++
−1
0 2
1ndx
x1x
1n1 ...(2)
from (1) ∫ +
+1
0 2
1ndx
x1x =
4π – (n + 1) In
so ∫ +
−1
0 2
1n
x1x dx =
4π – (n – 1)In–2
use it in (2)
In = )1n(4 +
π – n)1n(
1+
+ 1n
1+
−−
π−2nI)1n(
4
(n + 1) In + (n – 1) In – 2 = 2π –
n1
7. diff. partially w. r.t. x
f ´
+
3yx .
31 =
3)x´(f
Let x = 0 f ´(y/3) = f ´ (0) so f ´(x) = f ´(0) hence f ´(x) is a constant so f ´ (x) = 2 (as given) so f(x) = 2x + c; since f(0) = 2 so f(x) = 2x + 2 Hence f(2) = 6
8,9,10. g (f(x)); z → z1 g(f(x)) = 1 ; x ∈ even integer = 0 ; otherwise Hence g(f(x)) is many one onto, periodic and even
function Now, f(g(x)) = p ; if x is perfect square = 1 ; otherwise Hence f(g(x)) = p has infinitely many roots.
Do you know
• The largest telescope in the world is currently being constructed in northern Chile. The telescope will utilize four - 26 ft. 8 in. (8.13 meters) mirrors which will gather as much light as a single 52 ft. 6 in. (16 meters) mirror.
• The Hubble Space Telescope weighs 12 tons (10,896 kilograms), is 43 feet (13.1 meters) long, and cost $2.1 billion to originally build.
• The longest living cells in the body are brain cells which can live an entire lifetime.
• The largest flying animal was the pterosaur which lived 70 million years ago. This reptile had a wing span of 36-39 feet (11-11.9 meters) and weighed 190-250 pounds (86-113.5 kilograms).
• The Atlantic Giant Squid's eye can be as large as 15.75 inches (40 centimeters) wide.
• Armadillos, opossums, and sloth's spend about 80% of their lives sleeping.
• The starfish species, Porcellanaster ivanovi, has been found to live in water as deep as 24,881 feet (7,584 meters).
• The tentacles of the giant Arctic jellyfish can reach 120 feet (36.6 meters) in length.
• The greatest tide change on earth occurs in the Bay of Fundy. The difference between low tide and high tide can be as great as 54 ft. 6 in. (16.6 meters).
• The highest temperature produced in a laboratory was 920,000,000 F (511,000,000 C) at the Tokamak Fusion Test Reactor in Princeton, NJ, USA.
• The fastest computer in the world is the CRAY Y-MP C90 supercomputer. It has two gigabytes of central memory and 16 parallel central processor units.
XtraEdge for IIT-JEE 41 APRIL 2010
1. In = ∫ −a0
nx dxxe = ( )a0nx xe− + n ∫ −−a0
1nx xe dx = – e–a an + nIn – 1
In = –e–a an + n[–e–a an–1 + (n–1)In–2] = –e–a[an + nan–1 +n(n – 1)an–2 + n(n – 1)(n – 2)an–3 + ..... + n(n – 1)..... 2a] + n I0
= ( )
+++−
+−−−
−− a2
a....1n
an
aeen21nn
aa0
x
In =
++++− −
na.....
2aa1e1n
n2a
Now, I = ∫∞ −0
nx xe = ∞→a
Lt In = n
2. Mid pt. M of BC = (0, –1)
A(6, 5)
B(2, –3)
M
C (–2, 1) 60º
45º
P
slope of BC = 2213
+−− = –1
so slope of altitude of ∆PBC is = 1. length BC = 1616 + = 24
Now altitude PM = 24 sin 60º
= 24 . 23 = 62
eqn. of PM line is
21
0x − =
21
1y + = r (as its slope is 1)
x = 2r & y =
2r – 1
for req. pt. P take r = 62 .
So pt. P ≡
−1
262,
262 = )132,32( −
3. Sn = ∑=
n
0rr . nCr = n . 2n–1
so S = ∑=
n
1n1S = ∑
=
−n
1n
1n2.n
S = 1 + 2.21 + 3.22 + 4.23 + .... + n . 2n–1 2S = 2 + 2.22 + 3.23 + ..... + (n – 1). 2n–1 + n.2n (1 – 2)S = (1 + 2 + 22 + 23 + .... + 2n–1) – n . 2n
= 1 . 1212n
−− – n . 2n
S = n . 2n – 2n + 1 = (n – 1) 2n + 1 Now A.M. ≥ G. M.
n.....321
C.n....C.3C.2C nn
3n
2n
1n
++++++++
≥
( ) ( ) ( ) n...211
nn
n33
n22
n1
n C......C.C.C +++
2)1n(n
1n
2)1n(n
2.n
+
−
+ ≥ nC1 (nC2)2 ..... (nCn)n
so nC1 . (nC2)2 ..... (nCn)n ≤ 2
1n Cn
1n2
+
+
4. Let P be a point represented by 1. so as |z1 – 1| = |z2 – 1| = |z3 – 1| so P is the
circumcentre of ABC. Its centroid is 3
zzz 321 ++
If the ∆ABC is equilateral then circumcentre = centroid
so 1 = 3
zzz 321 ++
so z1 + z2 + z3 = 3 Now if z1 + z2 + z3 = 3 then centroid of ∆ABC is 1
which is point P and P is already the circumcentre of ∆ABC. So now if they are same then ∆ABC is equilateral.
5. let the centroid of ∆ABC be (h, k) then 3h = r cos α + r cos β + r
rh3 – 1 = 2 cos
2β+α cos
2β−α ...(1)
& rk3 = sin α + sin β
= 2 cos 2
β−α sin 2
β+α ...(2)
MATHEMATICAL CHALLENGES SOLUTION FOR THIS ISSUE (SET # 12)
XtraEdge for IIT-JEE 42 APRIL 2010
line BC, x cos2
β+α + y sin2
β+α = r cos2
β−α
Let point D be (a, b)
D C(α)
A(r, 0)
B(β)
then a cos 2
β+α + b sin 2
β+α = r cos 2
β−α
....(3)
Multiply (3) by cos 2
β−α
a cos2
β−α cos2
β+α + b sin2
β+α cos2
β−α
= r cos2
2β−α
use (1) & (2)
−1
rh3
2a + b
r2k3 = r cos2
2β−α ...(4)
square & add (1) & (2)
4 cos2
2β−α =
2
1rh3
− + 2
2
rk9 ...(5)
from (4) & (5)
2
1rh3
41
− + 2
2
r4k9 =
−1
rh3
r2a + 2r2
k3b
so req. locus is (3x – r)2 + 9y2 = 2a(3x – r) + 6.b.y 9x2 + 9y2 – 6rx + r2 = 6ax – 2ar + 6.b.y
x2 + y2 – 32 (r + a)x –
32 by +
9r2
+ 9ar2 = 0
22
b31y)ar(
31x
−+
+−
= 9
bar2r)ar( 222 +−−+ = 9
ba 22 +
It is a circle and radius is 3
ba 22 +
Since point D is interior of circle so a2 + b2 < r2, so radius of this circle is less than r/3. 6. Prime factors dividing 10 are 2, 3, 5, 7. As required
the number chosen should not be divisible by 2 or 3 or 5 or 7. Define, the events as
A : divisible by 2 B : divisible by 3 C : divisible by 5 D : divisible by 7
A ∪ B ∪ C ∪ D = A + B + C + D – A ∩ B – A ∩ C – A ∩ D – B ∩ C – B ∩ D – C ∩ D + A ∩ B ∩ C
+ A ∩ C ∩ D + B ∩ C ∩ D + A ∩ B ∩ D – A ∩ B ∩ C ∩ D
P(A ∪ B ∪ C ∪ D)
= 2006
1 [ 1003 + 668 + 401 + 286 – 334 – 200
– 143 – 133 – 95 – 57 + 66 + 28 + 19 + 47 – 9]
= 20061547
7. Line BC
Aa
H
DB Cb = –i + 3j c = 2i + 5j
→r = – j3i + + )j2i3(t + ...(1) any pt. D on it = (3t – 1) i + (3 + 2t) j As HD ⊥ BC, so ((3t – 1 – 1) i + (3 + 2t – 2) j). (3i + 2j) = 0 3(3t – 2) + 2(2t + 1) = 0
13 t – 4 = 0 ⇒ t = 134
so point D = –13i +
13j47
Now line HD ⇒ →r = i + 2j + s´
−−+− j2i
13j47
13i
= i + 2j + s(–14i + 21j)
⇒ →r = i + 2j + λ(–2i + 3j) Any point A on it = (1 – 2λ)i + (2 + 3λ)j Now as AC ⊥ BH so [ (1 – 2λ – 2)i + (2 + 3λ – 5)j] . [2i – j] = 0 2 (–1 – 2λ) – (3λ – 3) = 0
⇒ –7λ = 2 – 3 ⇒ λ = 71
so pt. A = 7i5 +
7j17
8. If A is the area of the triangle with sides a, b, and c,
then A2 = s(s – a) (s – b) (s – c); where 2s = a + b + c using AM – GM inequality for s – a, s – b, s – c, we
have
A2 ≤ s3
3)cs()bs()as(
−+−+−
XtraEdge for IIT-JEE 43 APRIL 2010
A2 ≤ s 3
3s2s3
− = 3
4
3s
A ≤ 33
s2
Let 2s = p, then A ≤ 312
p2
Amax = 312
p2,
As condition of equality holds iff s – a = s – b = s – c which happen if a = b = c so Amax = ; for a = b = c
Now again p ≥ A312
pmin. = A312 ; and again equality holds if a = b = c. 9. Let the amount of salt dissolved at any time t is x kg.
So concentration is 300
x
so dtdx = k
−
300x
31 = k
−
300x100
x100
dx−
= 300k dt
– ln (100 – x) = 300k t + C
at t = 0, x = 0 so C = –ln 100
so 300kt = ln 100 – ln (100 – x) = ln
x100100
−
at t = 1 min., x = 1/3
so 300k = ln
31100
100
−
so 300k = ln
3299100
so 300k = ln
299300
so k = 300 ln 299300
so ln 299300 . t = ln
x100100
−
so t
299300
=
x100100
−
so 100 – x = 100 . t
300299
so x = 100
−
t
3002991
at t = 60 min.
x = 100
−
60
3002991 kg
10. Let P (3 cos θ, 3 sin θ)
line BC : y = – 3 ; line AC : A
B C
P
3x cos θ +
3y sin θ = 1
pt. C
−
θθ+ 3,
cos)sin1(3
pt. A (0, 3 cosec θ )
Area A = 21 . 2 .
θθ+
cos)sin1(3 . ( 3 cosec θ + 3 )
= θθθ+
cossin)sin1(33 2
= θ
θ+2sin
)sin1(36 2
θddA =
θθθ+−θθθ+
2sin)2cos)sin1(2cos2sin)sin1(2(36
2
2
= θ
θθ−θ−θθθ+2sin
)2cossin2coscos2)(sinsin1(3122
= θ
θ−θ−θθ+2sin
)2cos)2)(sin(sin1(3122
= θ
θ+−θθ+2sin
)sin21)(sinsin1(3122
2
= θ
+θ−θθ+2sin
)1)(sin1sin2)(sin1(3122
= θ
−θθ+2sin
)1sin2()sin1(3122
2
Amin at θ = 6π
so Amin =
23
21136
2
+
= 12 . 49
= 27sq. units. = 0027 Ans.
XtraEdge for IIT-JEE 44 APRIL 2010
1. Let f : R → R and f(x) = g(x) + h(x), where g(x) is a
polynomial and h(x) is a continuous and differentiable bounded function on both sides, then f(x) is onto if g(x) of odd degree and f(x) is into if g(x) is of even degree. Then check whether f(x) is one one, many one, onto or into.
(i) f(x) = a1x + a3x3 + a5x5 + .... + a2n + 1x2n + 1 – cot–1x where 0 < a1 < a2 < a3 ....... < a2n + 1
(ii) f(x) = 1xx
2x)1x)(1x(x2
44
++
++++
Sol. (i) f(x) = odd degree polynomial + bounded function cot–1x also f´(x) > 0
⇒ y = f(x) will be one one and onto
(ii) f(x) = 1xx
)1xx)(1x(2
24
+++++ = x4 + 1 +
1xx1
2 ++
= even degree polynomial
+ bounded function ∈
34,0
also f´(x) = 0 has at least one root real which is not repeated since f´(x) is a polynomial of degree 7.
⇒ f(x) = 0 has at least one point of extrema. ⇒ many one & Into 2. A rectangle ABCD of dimensions r and 2r is folded
along the diagonal BD such that planes ABD and CBD are perpendicular to each other. Let the position of the vertex A remains unchanged and C0 is the new position of C, then find the distance of C0 from A and shortest distance between the edges AB & C0D.
(2r, r)
C
B (2r, 0)(0, 0) A
D (0, r)
N
Sol. Let the rectangle ABCD lies on the plane xy. After
folding the rectangle along the BD co-ordinates of points in 3-D are-
A : (0, 0, 0), B : (2r, 0, 0), C : (2r, r, 0), D(0, r, 0)
and N :
0,
5r,
5r2 and C0
5r2,
5r,
5r2 ,
Now AC0 = 585 r
and shortest distance = |DCAB|
)DCAB.(AC
0
00
×
×=
3r5 unit.
3. Let f(x) is a polynomial one-one function such that
f(x).f(y) + 2 = f(x) + f(y) + f(xy) ∀ x, y ∈ R–0,
f(1) ≠ 1, f´(1) = 3. Let g(x) = 4x (f(x) + 3) – ∫
x
0)x(f dx,
then prove that g(x) is an identity for all given x ∈ R –0.
Sol. putting x = y = 1 in given condition we get f(1)2 + 2 = f(1) + f(1) + f(1) ⇒ f(1)2 – 3f(1) + 2 = 0 ⇒ f(1) = 1 or 2 ⇒ f(1) = 2 Now put y = 1/x,
f(x) . f
x1 + 2 = f(x) + f
x1 + f(1)
⇒ f(x) = 1 ± xn According to given conditions, f(x) = 1 + x3
Now, g(x) = 4x [1 + x3 + 3] – ∫
x
0(1 + x3) dx = 0
⇒ g(x) = 0 for ∀ x ∈ R –0 4. Let three normals are drawn to the parabola y2 = 4ax
at three points P, Q and R, from a fixed point A. Two circles S1 and S2 are drawn on AP and AQ as diameter. If slope of the common chord of the circles S1 and S2 be m1 and the slope of the tangent to the parabola at the point R be m2, then prove that m1 . m2 = 2.
Sol. Let A(h, k) be a fixed point at3 + (2a – h)t – k = 0 Q three normals are drawn from (h, k) Let feet of normals P,Q and R are three points with
parameters t1, t2 and t3. Common chord of S1 and S2 = S1 – S2 = (t1 + t2)x +
2y – h(t1 + t2) – 2k = 0 Tangent to the parabola at R = t3y = x + at3
2
m1 . m2 = –
+
2tt 21 .
3t1 =
21
(Q t1 + t2 + t3 = 0 for co-normal points).
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
MATHS
XtraEdge for IIT-JEE 45 APRIL 2010
5. Let z1, z2 and z3 are unimodular complex numbers then find the greatest value of |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2.
Sol. |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2 = 2[|z1|2 + |z2|2 + |z3|2] – [z1 2z + 1z z2 + 1z z3 + 3z z1 + z2 3z + 2z z3] = 6 – [z1 2z + z2 1z + z1 3z + z3 1z +z2 3z + z3 2z ] ...(1) Now |z1 + z2 + z3|2 ≥ 0 ⇒ z1 2z + z2 3z + z2 1z + z3 2z + z1 3z + 1z z3 ≥ –3 ...(2) From (1) & (2) maxm value of |z1 – z2|2 + |z2 – z3|2 + |z3 – z1|2 = 6 – (–3) = 9 6. Consider following two infinite series in real θ and r
C = 1 + r cos θ + !2
2cosr 22 θ + !3
3cosr3 θ + ....
S = r sin θ + !2
2sinr2 θ + !3
3sinr3 θ + .....
If a remains constant and r varies the prove that
(i) CdrdC + S
drdS = (C2 + S2) cos θ
(ii) 2
drdC
+
2
drdS
= C2 + S2
Sol. We have C + iS =
θiree ...(1)
C – iS = θ−iree ...(2)
Now,
C2 + S2 = 2
reie
θ =
2sinricosr e.e θθ = e2r cos θ ...(3)
Differentiating (1) w.r.t.r, we get
drdC + i
drdS = eiθ .
θie.re ....(4)
∴ 2
drdC
+
2
drdS
=
θθ ie.ri e.e = e2r cos θ
= C2 + S2 (Form 3) multiply (2) and (4)
+
drdSi
drdC (C – iS) = eiθ .
θie.re .θ−ie.re
= eiθ ]e[ )ee(r ii θ−θ + = eiθ . e2r cosθ Now equating real parts in both sides
CdrdC + S
drdS = (C2 + S2) cosθ
Hence proved.
Science Facts
Skin Deep Storage
Chip implants that keep track of personal information seem like a novelty but do they have a more useful future?
These days, some people are following their pets and getting tagged. Radio frequency identification (RFID) chips are the size of a grain of rice and can be loaded up with personal information like passwords and implanted under the skin. Instead of having to remember a login code, an RFID reader can be set up to automatically detect it and grant you access to a range of things from your computer to your front door. It seems like it could be useful to people with exceptionally poor memories, but right now these chips are being snapped up by technology geeks like Amal Graafstra. The 29-year-old businessman from Vancouver, Canada, is one of the first people to have an RFID implant and so far is happy with the results. "I just don't want to be without access to the things that I need to get access to. In his chip, he has stored a unique identification number which can be used to log him into various electronic devices. It didn't cost him an arm and a leg either: he got the whole set-up on the internet for about $50 (£30), including the $2 cost of the chip itself. The procedure to implant the chip is quite simple and painless. Amal's chip was implanted under the skin of his left hand while he was under a local anesthetic. It is possible to inject the chips using a large enough needle, but in Amal's case the chip was inserted by simply cutting through his skin with a scalpel. Other than complaining of sensitivity in the area of the implant, Amal said that it doesn't hurt and he expects that eventually the chip will be completely unobtrusive. A hand implanted with an RFID chip and the chip reader. The chip is made of silicon and is digitally encoded with information.. A RFID reader, which is installed in a computer or an electronic device like a reader by a front door, emits a radio signal of a particular frequency, just like radio stations each broadcast on their own frequency. The chip acts passively when it is within 3 inches of the reader: the right incoming radio signal induces just enough energy in the antenna of the chip for a circuit in the chip to power up and produce a response. The reader can then access the information on the chip and pass it on to the computer or device that requires it.
XtraEdge for IIT-JEE 46 APRIL 2010
Function, Limits, Continuity & Differentiability :
If the domain of the function is in one quadrant then the trigonometrical functions are always one-one.
If trigonometrical function changes its sign in two consecutive quadrants then it is one-one but if it does not change the sign then it is many one.
In three consecutive quadrants tigonometrical functions are always many one.
Any continuous function f(x), which has at least one local maximum, is many-one.
Any polynomial function f : R → R is onto if degree of f is odd and into if degree of f is even.
An into function can be made onto by redefining the cordomain as the range of f is even.
An into function can be made onto by redefining the codomain as the range of the original function.
If f(x) is periodic with period T then )x(f
1 is also
periodic with same period T.
If f(x) is periodic with period T, )x(f is also periodic with same period T.
Period of x – [x] is 1. Period of algebraic functions x , x2, x3 + 5 etc. does't exist.
If )x(flimax→
does not exist, then we can not remove
this discontinuity. So this become a non-removable discontinuity or essential discontinuity.
If f is continuous at x = c and g is discontinuous at x = c, then
(a) f + g and f – g are discontinuous (b) f.g may be continuous
If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous.
Point functions (domain and range consists one value only) is not a continuous function.
If a function is differentiable at a point, then it is continuous also at that point.
i.e., Differentiability ⇒ Continuity, but the converse need not be true.
If a function 'f' is not differentiable but is continuous at x = a, it geometrically implies a sharp corner or kink at x = a.
If f(x) and g(x) both are not differentiable at x = a then the product function f(x).g(x) can still be differentiable at x = a.
If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x) + g(x) is also not differentiable at x = a.
If f(x) and g(x) both are not differentiable at x = a, then the sum function may be a differentiable function.
Differentiation and Applications of Derivatives :
dxdy is
dxd (y) in which
dxd is simply a symbol of
operation and not 'd' divided by dx. If f´(x0) = ∞, the function is said to have an infinite
derivative at the point x0. In this case the line tangent to the curve of y = f(x) at the point x0 is perpendicular to the x-axis.
Of all rectangles of a given perimeter, the square has the largest area.
All rectangles of a given area, the squares has the least perimeter.
A cone of maximum volume that can be inscribed in
a sphere of a given radius r, is of height 3r4 .
A right circular cylinder of maximum volume that can be inscribed in a square of radius r, is of height
3r2 .
If at any point P(x1, y1) on the curve y = f(x), the tangent makes equal angle with the axes, then at the
point P, ψ = 4π or
43π . Hence, at P tan ψ =
dxdy = ±1
Indefinite Integral : If F1(x) and F2(x) are two antiderivatives of a
function f(x) on an interval [a, b], then the difference between them is a constant.
The signum function has an antiderivative on any interval which does not contain the point x = 0, and does not possess an anti=derivative on any interval which contains the point.
The antiderivative of every odd function is an even function and vice-versa.
CALCULUS Mathematics Fundamentals M
AT
HS
XtraEdge for IIT-JEE 47 APRIL 2010
If In = ∫ axn e.x dx, then In = aex axn
– an In–1
If In = ∫ dx)x(log , then In = x log x – x
If In = ∫ xlog1 dx, then
In = log(logx) + logx + )!2.(2)x(log 2
+ )!3.(3)x(log 3
+ ...
If In = ∫ dx)x(log n ; then In = x(logx)n – n.In–1
Successive integration by parts can be performed when one of the functions is xn (n is positive integer) which will be successively differentiated and the other is either of the following sin ax, cos ax, e–ax, (x +a)m which will be successively integrated.
Chain rule :
∫ dxv.u = uv1 – u´v2 + u"v3 – u"'v4 + ....
+ (–1)n – 1un–1vn + (–1)n ∫ dxv.u n
where un stands for nth differential coefficient of u and vn stands for nth integral of v.
∫ axxe sin(bx + c)dx = 22
ax
baxe
+[a sin(bx + c) – b
cos(bx + c)] – 222
ax
)ba(e+
[(a2 – b2)sin (bx + c) – 2ab
cos (bx + c)] + k
∫ axxe cos(bx + c)dx = 22
ax
bae.x+
[a cos(bx + c) – b
sin(bx + c)] – 222
ax
)ba(e+
[(a2 – b2)cos (bx + c) – 2ab
sin (bx + c)] + k
∫ axxe sin(bx + c)dx
= 22
x
b)a(loga
+[(loga)sin(bx + c) – b cos(bx + c)] + k
∫ axxe cos(bx + c)dx
= 22
x
b)a(loga
+[(loga)cos(bx + c) + b sin(bx + c)] + k
∫ ++
xsindxcoscxsinbxcosa dx
= 22 dcbdac
++ x + 22 dc
bcad+− log |c cos x + d sinx| + k.
Reduction formulae for I(n,m) = ∫ xcosxsin
m
n dx is
I(n,m) = 1m
1−
.xcosxsin
1m
1n
−
−
– )1–m()1n( − .I(n–2, m – 2)
Definite Integral and Area Under Curves :
The number f(c) = ∫−
b
adx)x(f
)ab(1 is called the
mean value of the function f(x) on the interval [a, b]. If m and M are the smallest and greatest values of a
function f(x) on an interval [a, b], then m(b – a) ≤
∫b
adx)x(f ≤ M(b – a).
If f2(x) and g2(x) are integrable on [a, b], then
∫b
adx)x(g)x(f ≤
2/1b
a
2 dx)x(f
∫
2/1b
a
2 dx)x(g
∫
Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t1), b = φ(t2), then
∫b
adx)x(f = ∫ φφ
2
1
t
tdt)t´()t((f
Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈[c, d], if I(α) =
∫ αb
adx),x(f , then I´(α) = ∫ α
b
adx),x´(f , where
I´(α) is the derivative of I(α) w.r.t. α and f´(x, α) is the derivative of f(x, α) w.r.t α, keeping x constant.
∫b
adx)x´(f = (b – a) ∫ +−
1
0dt]at)ab[(f
∫ −++
b
a )xba(f)x(fdx)x(f =
21 (b – a)
The area of the region bounded by y2 = 4ax, x2 = 4by
is 3ab16 sq. unit.
The area of the region bounded by y2 = 4ax and y =
mx is 3
2
m3a8 sq. unit.
The area of the region bounded by y2 = 4ax and its
latus-rectum is 3a8 2
sq. unit.
The area of the region bounded by one arch of sin(ax) or cos (ax) and x-axis is 2/sq. unit.
Area of the ellipse (x2/a2) + (y2/b2) = 1 is πab sq. unit. Area of region bounded by the curve y = sin x, x-axis
and the line x = 0 and x = 2π is 4 sq. unit.
XtraEdge for IIT-JEE 48 APRIL 2010
Complex Numbers :
|z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |1m(z) | ≥ 1m(z)
|z|
z is always a unimodular complex number if z ≠ 0
|Re(z) | + |1m(z) | ≤ 2 |z|
If z1z + = a, the greatest and least values of |z| are
respectively 2
4aa 2 ++ and 2
4aa 2 ++−
|z1 + 22
21 zz − | + |z2 – 2
221 zz − | = |z1 + z2| + |z1 – z2|
If z1 = z2 ⇔ |z1| = |z2| or arg z1 = arg z2 |z1 + z2| = |z1 – z2| ⇔ arg (z1) – arg(z2) = π/2. If |z1| ≤ 1, |z2| ≤ 1 then
(i) |z1 + z2|2 ≤ (|z1| – |z2|)2 + (arg (z1) – arg (z2))2 (ii) |z1 + z2|2 ≥ (|z1| + |z2|)2 – (arg (z1) – arg(z2))2
|z1 + z2|2 = |z1|2 + |z2|2 + 2||z2| cos(θ1 – θ2). |z1 – z2|2 = |z1|2 + |z2|2 – 2|z1||z2| cos(θ1 – θ2). If z1 and z2 are two complex numbers then
|z1 z2| = r1r2; arg(z1z2) = θ1 + θ2 and 2
1
zz =
2
1
rr ,
arg
2
1
zz = θ1 – θ2 and where |z1| = r1, |z2| = r2,
arg(z1) = θ1 and arg(z2) = θ2. The area of the triangle whose vertices are z, iz and
z + iz is 21 |z|2.
The area of the triangle with vertices z, wz and
z + wz is 43 |z2|.
If z1, z2, z3 be the vertices of an equilateral triangle and z0 be the circumcentre, then
21z + 2
2z + 23z = 3 2
0z . If z1, z2, z3 ....... zn be the vertices of a regular
polygon of n sides and z0 be its centroid, then 2
1z + 22z + .......+ 2
nz = n 20z .
If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral if
(z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0 or 2
1z + 22z + 2
3z = z1z2 + z2z3 + z3z1
or 21 zz
1−
+ 32 zz
1−
+ 13 zz
1−
= 0
If z1, z2 z3 are the vertices of an isosceles triangle, right angled at z2 then 2
1z + 22z + 2
3z = 2z2(z1 + z2) If z1, z2, z3 are the vertices of a right-angled isosceles
triangle, then (z1 – z2)2 = 2(z1 – z3)(z3 – z2). If z1, z2, z3 be the affixes of the vertices A, B, C
respectively of a triangle ABC, then its orthocentre is
CseccBsecbAseca
z)Csecc(z)B(secbz)A(seca 321
++++
For any a, b ∈ R
(i) iba + + iba − = aba2 22 ++
(ii) iba + – iba − = aba2 22 −+
The sum and product of two complex numbers are real simultaneously if and only if they are conjugate to each other.
If ω and ω2 are the complex cube roots of unity, then (i) (aω + bω2)(aω2 + bω) = a2 + b2 – ab (ii) (a + b (aω + bω2)(aω2 + b2ω) = a3 + b3 (iii) (a + bω + cω2)(a + bω2 + cω) = a2 + b2 + c2 – ab – bc – ca (iv) (a + b + c) (a + bω + cω2) (a + bω2 + cω) = a3 + b3 + c3 – 3abc If three points z1, z2, z3 connected by relation
az1 + bz2 + cz3 = 0 where a + b + c = 0, then the three points are collinear.
If three complex numbers are in A.P. then they lie on a straight line in the complex plane.
Progression : If Tk and Tp of any A.P. are given, then formula for
obtaining Tn is knTT kn
−− =
kpTT kp
−
−.
If pTp = qTq of an A.P., then Tp + q = 0. If pth term of an A.P. is q and the qth term is p, then
Tp+q = 0 and Tn = p + q – n. If the pth term of an A.P. is 1/q and the qth term is 1/p,
then its pqth term is 1.
ALGEBRA Mathematics Fundamentals M
AT
HS
XtraEdge for IIT-JEE 49 APRIL 2010
The common difference of an A.P. is given by d = S2 – 2S1 where S2 is the sum of first two terms and S1 is the sum of first term or the first term.
If sum of n terms Sn is given then general term Tn = Sn – Sn–1, where Sn–1 is sum of (n – 1) terms of A.P.
If for an A.P. sum of p term is q and sum of q terms is p, then sum of (p + q) terms is –(p + q).
If for an A.P., sum of p term is equal to sum of q terms, then sum of (p + q) terms is zero.
If the pth term of an A.P. is 1/q and qth term is 1/p,
then sum of pq terms is given by Spq = 21 (pq + 1).
Sum of n A.M.'s between a and b is equal to n times the single A.M. between a and b.
i.e. A1 + A2 + A3 + ...... + An = n
+
2ba .
If Tk and Tp of any G.P. are given, then formula for
obtaining Tn is kn
1
k
n
TT −
=
kp1
k
p
TT −
Product of n G.M.'s between a and b is equal to nth power of single geometric mean between a and b
i.e. G1G2G3 .... Gn = ( ab )n. The product of n geometric means between a and 1/a
is 1.
If n G.M.'s inserted between a and b then r = 1n
1
ab +
Quadratic Equations and Inequations : An equation of degree n has n roots, real or
imaginary. If f(α) = 0 and f´(α) = 0, then α is a repeated root of
the quadratic equation f(x) = 0 and f(x) = a(x – α)2. In fact α = –b/2a.
If α is repeated common root of two quadratic equations f(x) = 0 and φ(x) = 0, then α is also a common root of the equations f´(x) = 0 and φ´(x) = 0.
In the equation ax2 + bx + = 0 [a, b, c ∈R], if a + b + c = 0 then the roots are 1, c/a and if a – b + c = 0, then the roots are –1 and – c/a.
If one root of the quadratic equation ax2 + bx + c = 0 is equal to the nth power of the other, then
( ) 1n1
nac + + ( ) 1n1
nca + + b = 0. If one root is k times the other root of the quadratic
equation ax2 + bx + c = 0, then k
)1k( 2+ =acb2
.
If an equation has only one change of sign, it has one +ve root and no more.
Permutations and Combinations : nC0 = nCn = 1, nC1 = n nCr + nCr–1 = n+1Cr nCx = nCy ⇔ x = y or x + y = n n. n-1Cr–1 = (n – r + 1)nCr–1 If n is even then the greatest value of nCr is nCn/2.
If n is odd then the greatest value of nCr is 2
C 1nn
+ or
2C 1n
n− .
nCr = rn .n–1Cr–1.
Number of selection of zero or more things out of n different things is, nC0 + nC1 + bC2 + ... + nCn = 2n.
nC0 + nC2 + nC4 + .... = nC1 + nC3 + nC5 + .... = 2n–1. Gap method : Suppose 5 moles A, B, C, D, E are
arranged in a row as × A × B × C × D × E ×. There will be six gaps between these five. Four in between and two at either end. Now if three females P, Q, R are to be arranged so that no two are together we shall use gap method i.e., arrange them in between these 6 gaps. Hence the answer will be 6P3.
Together : Suppose we have to arrange 5 persons in a row which can be done in 5! = 120 ways. But if two particular persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 – 2 + 1 (1 corresponding to these two together) = 3 + 1 = 4 units, which can be arranged in 4! ways. Now we loosen the string and these two particular can be arranged in 2! ways. Thus total arrangements = 24 × 2 = 48.
If we are given n different digits (a, a2, a3 ..... an) then sum of the digits in the unit place of all numbers formed without repetition is (n – 1)!(a1 + a2 + a3 + .... + an). Sum of the total numbers in this case can be obtatined by applying the formula (n – 1)!(a1 + a2 + a3 + ..... + an). (1111 ......... n times).
Binomial Theorem & Mathematical Induction : The number of terms in the expansion of (x + y)n are
(n + 1). If the coefficients of pth, qth terms in the expansion of
(1 – x)n are equal, then p + q = n + 2. For finding the greatest term in the expansion of
(x + y)n. we rewrite the expansion in this form
(x + y)n = xnn
xy1
+ . Greatest term in (x + y)n = xn.
Greatest term in n
xy1
+ .
There are infinite number of terms in the expansion of (1 +x)n, when n is a negative integer or a fraction.
The number of term in the expansion of (x1 + x2 + .... + x2)n = n+r-1Cr–1.
XtraEdge for IIT-JEE 50 APRIL 2010
Learning the MIghTy way
When you hear... you forget. When you see... you remember. When you do... you understand. – Confucius
This sounds very much the way we learnt things in our childhood. Our parents did not tell us how to walk, but made us walk. They did not show us how to ride a bicycle, they made us ride it. And that’s how learning goes naturally: deriving inspiration by doing things, ‘Inspired Learning’.
In this era of information overload, educational institutions hardly devote time and resources to make students follow this natural path of learning. An exception to this is Manipal Institute of Technology (MIT), where the ‘been there, done that’ spirit has helped students make a mark worldwide.
While in other colleges, students were seeing how a car is built, students at MIT designed and manufactured a high performance car which made its way into the ‘Formula Student 2009’ (Student edition of Formula One), at Silverstone, UK. MIT got 65th position among teams from 126 universities of 23 countries. Several other feathers of innovation have been added to MIT’s achievement hat within the last 2 years like: 1st position in “Train Blazer- 2008”; 1st position in “GE Edison Innovation Challenge -2008”; 1st and 2nd position in “Schneider Electric Innovation Challenge” 2008; 1st prize for an Innovative proposal on “Solar Power embedded Street Lamp” and many more.
The foundation of this hub of thinking and innovation was laid 52 years ago, and since then is being strengthened by constructive effort and hard work. The quality of technical education is evident in the research activities going on in the campus: 6 patents filed in the current year, 9 grants received for various activities and, several papers, books and awards added to its research portfolio. A major landmark is ‘Manipal University Technology Business Incubator’, which is being established with a huge funding from the Department of Science and Technology, Government of India.
While focusing on innovation, MIT has taken care of students by getting them absorbed into the best industries, through quality industry projects and collaborations with universities and companies. ‘Practice School’ is one such concept, in sync with the idea of ‘Inspired learning’. This was introduced at MIT in 2005, through which students are trained to effectively link the theory learnt in classrooms with the practice in the industry. This training helps reduce the cost for a company, on in-house training of students who are their prospective employees. Due to such efforts placement statistics have been moving up since 2005 reaching 95% in 2008. Even in the recession hit 2008-09, 56 companies turned up offering jobs to 948 students, with the best remuneration reaching 10.75 lakhs per annum. MIT now stands 7th among the private Engineering colleges in India, and 22nd overall.
A student, Gaurav Sinha (4th year, Electrical and Electronics) quotes “MIT has provided me with excellent opportunities to develop to my full potential. The environment is conducive for our growth and along with the state-of-art infrastructure facilities available helps bring out the best in each of us. The platform and exposure that I got at MIT has groomed my personality and prepared me to face any challenge in future”. Already having produced strong personalities of industry like Rajeev Chandrasekhar (Chairman & CEO of Jupiter Capital; FICCI President; Founder & former-CEO of BPL mobile), Ravi Bapna (Executive Director of CITNE & Professor, Indian School of Business) , Amit Behl (Director, Intel India), etc. the excellent infrastructure coupled with innovative learning and course content, holds promise for a much brighter future for the country.
– MIT, Manipal
XtraEdge for IIT-JEE APRIL 2010 51
XtraEdge for IIT-JEE APRIL 2010 52
CHEMISTRY
SECTION – I Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1. A bis-aldol dimerization of 1-phenyl-1,2-propanedione (C6H5COCOCH3) gives which of the following ?
(A)
O
O
C6H5
C6H5 (B)
OC6H5
C6H5
(C) O C6H5
C6H5 (D)
OC6H5
O C6H5
2. An alkene (A), C16H16 on ozonolysis gives only one product B(C8H8O). Compound (B) on reaction with NH2OH, H2SO4, ∆ gives N-methyl benzamide. The compound 'A', is –
(A) CH3 C = C
H CH3 H
(B)
C = C CH3 CH3
(C) CH2–CH = CH–CH2
(D) CH = CH
CH3
CH3
3. Identify product D in the following reaction sequence:
H3C–
3
3
CH|
—C|
CH
CH2CH2OHHeat,OH
H,OCrK2
722 →+
A →2SOCl B
→NH)CH( 23 COH.II
ether,LiAlH.I
2
4 → D
(A) H3C–
3
3
CH|
—C|
CH
CH2C ≡ N
(B) H3C–
3
3
CH|
—C|
CH
CH2 23
23
)CH(CHN|
)CH(N
(C) H3C –
3
3
CH|
—C|
CH
CH2 C||O
N(CH3)2
(D) H3C–
3
3
CH|
—C|
CH
CH2CH2N(CH3)2
MOCK TEST FOR IIT-JEE
PAPER - I
Time : 3 Hours Total Marks : 240
Instructions : • This question paper contains 60 questions in Chemistry (20), Mathematics (20) & Physics (20). • In section -I (8 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques) of each paper +4 marks will be given for correct answer –1 mark for wrong answer. • In section -III contains 2 groups of questions (2 × 3 = 6 Ques.) of each paper +4 marks will be given for each
correct answer & –1 mark for wrong answer. • In section -IV (2 Ques.) of each paper +8(2×4) marks will be given for correct answer & No Negative marking for
wrong answer.
XtraEdge for IIT-JEE APRIL 2010 53
4. Consider the following reactions : Reaction I :
H3C
O
O
O
CF3 1
23r
OHCHCH → H3C
O
O +
F3C
O
OH
Reaction II :
H3C
O
O
O
CF3 2
23r
OHCHCH →
H3C
O
OH +
F3C
O
O Which of the following is a correct comparison of the
rate of reactions I and II ? (A) rI = rII (B) r1 > rII (C) r1 < rII (D) None
5. A container containing HCl(g) is fitted with a funnel and a long capillary tube as shown below. Capillary is immersed in the water. When few drops of water is introduced into
the container through the funnel then – A
HCl(g)
(A) there is a pressure drop in the container and liquid
level in the capillary rises (B) there is a pressure drop in the container because
HCl(g) effuse out more rapidly than the air effuse in
(C) HCl(g) undergoes spontaneous dissociation to H2(g) & Cl2(g), hence number of moles decreases, resulting pressure drop inside the container. Water level in the capillary rises
(D) HCl(g) spontaneously mixed with water through the capillary therefore, water level in the capillary remain same
6. As per Boyle's law V ∝ 1/P at constant temperature,
As per charles law V ∝ T at constant pressure. Therefore, by combining, one concluded that T ∝ 1/P hence, PT = constant
(A) PT = constant is correct, because volume remain same in both the laws
(B) PT = constant is incorrect, because volume remain same at the constant temperature and at the constant pressure
(C) PT = constant is correct, because volume at constant temperaute and volume at constant pressure are not same
(D) PT = constant is incorrect, because volume at constant temperature and volume at constant pressure for the same amount of gas are different
7. Which of the following statements is correct ? (A) (n – 1) d subshell has lower energy than ns
subshell (B) (n – 1) d subshell has higher energy than ns
subshell (C) (n + 1) d subshell has lower energy than nf
subshell (D) nf subshell has lower energy than (n + 2) s
subshell
8. (A), (B) and (C) are elements in the third short period. Oxide of (A) is ionic , that of (B) is amphoteric and of (C) a giant molecule. (A), (B) and (C) have atomic numbers in the order of -
(A) (A) < (B) < (C) (B) (C) < (B) < (A) (C) (A) < (C) < (B) (D) (B) < (A) < (C)
SECTION – II Multiple Correct Answers Type
This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
9.
HH
+ Ph3C⊕BF4– →
⊕
H
BF4– + Ph3 CH
which of the following statements is/are correct -
(A) the cation in reactant side is approximately
1011 times more stable than product side
(B) cation in reaction side is non planer
(C) it is acid base reaction
(D) reaction must be exothermic
XtraEdge for IIT-JEE APRIL 2010 54
10. Which of the following is/are correct ? (A) The efficiency of a solid catalyst depends
upon its surface area (B) Catalyst operates by providing alternate path
for the reaction that involves lower activation energy
(C) Catalyst lowers the activation energy of forward reaction only without affecting the activation energy of backward reaction
(D) Catalyst does not affect the overall enthalpy change of the reaction
11. For the three elements P, Q & R, ionization
enthalpy (IE) and electron gain enthalpies (∆eg H) are given in the following table -
Element IE in kJ/mol ∆eg H in kJ/mol P 1680 –340 Q 1100 –120 R 500 –20
(A) P is the highest electronegative element among P, Q and R
(B) R is the least electronegative among P, Q & R (C) Electro negativity of P is approximately equal to 4 (D) R may be chlorine 12. Which of the following is/are correct ?
(A) For the incompressible liquid TdP
dH
is
approximately equal to volume of liquid
(B) For ideal gas TdP
dH
is equal to zero
(C) For real gas if TdV
dE
= 0 then not
necessarily TdP
dH
is equal to zero
(D) None of these
SECTION – III
Comprehension Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.
Paragraph # 1 (Ques. 13 to 15) A pleasant smelling optically active compound,
monoester 'F' has molecular weight 186. It doesn't react with Br2 in CCl4. Hydrolysis of 'F' gives two optically active compounds 'G', which is soluble in NaOH and 'H'. H gives a positive iodoform test, but on warming with conc. H2SO4 gives I with no disastereomers. When the Ag+ salt of 'G' is reacted with Br2, racemic 'J' is formed. Optically active J is formed when 'H' is treated with tosyl chloride (TsCl), and then with NaBr.
13. The pleasant smelling optically active compound, F is -
(A) (CH3)2CH–
3CH|
–CHC||O
O–
3CH|
–CH CH(CH3)2
(B) (CH3)3C–CH2 C||O
–O–
3CH|
–CH CH(CH3)2
(C) CH3CH2CH2
3CH|
–CH C||O
–O–
3CH|
–CH CH2CH2CH3
(D) CH3CH2
3CH|
–CH CH2– 2COCH||O
–
3
2
CH|
CHCH –CH3
14. How would be the structure of F if I exists as
diastereomers ?
(A) (CH3)2CH
3CH|
CHCO||O
3
23
CH|
)CH(CHCH
(B) (CH3)3CCH2 C||O
O
3
23
CH|
)CH(CHCH
(C) CH3CH2CH2
3CH|CHCO
||O
3
322
CH|
CHCHCHCH
(D) CH3CH2
3
2
HC|
CHCH C||O
OCH2
3
32
CH|
CHCHCH
XtraEdge for IIT-JEE APRIL 2010 55
15. What would be the structure of F if H gives negative idoform test ?
(A) (CH3)2
3
2
CH|
COCHCH||O
3
23
CH|
)CH(CHCH
(B) CH3(CH2)3CH2
3CH|
COCHCH||O
CH(CH3)2
(C) CH3CH2CH2
33
2
CHCH||
CHCOCHCH||O
CH2CH3
(D) CH3CH2
3
2
CH|
COCHCH||O
CH2
3
2
CH|
CHCH CH3
Paragraph # 2 (Ques. 16 to 18) To the 100 ml of 10–2 (M) aqueous solution of HCl
0.1 (M) HA (Ka = 10–2) is added in such a way so that the final pH of the solution become 1.7
Given log 2 = 0.3 16. What was the pH of 10–2 (M) aqueous solution of
HCl ? (A) pH = 2 (B) pH < 2 (C) pH > 2 (D) pH = 1.7 17. What volume of 0.1 (M) HA was required to add in
aqueous HCl to reduced the final pH equal to 1.7 ? (A) 175 ml (B) 100 ml (C) 104 ml (D) 75 ml 18. Which of the following solution is isohydric with
0.1(M) aqueous solution of HA ? (A) 0.01(M) aqueous solution of HB (Ka = 10–2) (B) 0.01(M) aqueous solution of HC (Ka=2×10–1) (C) 0.01(M) aqueous solution of HNO3 (D) 1(M) aqueous solution of HD (Ka = 10–3)
SECTION – IV
Matrix – Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
19. Match the following : Column-I Column-II (A)CH3CH2CH2NH2 (p)Treatment of NaNO2, HCl gives N-nitroso compound (B) CH3CH2NHCH3 (q)Treatment of NaNO2, HCl gives diazonium chloride
(C)CH3–
3CH|N − CH3 (r) Treatment of
CH3I (excess) followed by AgOH; heat gives out alkene
(D) NH2 (s) Treatment of HCl,
heat gives dealkylation (t) Treatment with CHCl3/
Alc.KOH gives isocyanide 20. Match the following : Column-I Column-II
(A) C6H5CH2CHO →− )eq1(Br,OH 2 (p) Redox products
(B) CH3CHO )Cº50(
OH
− →
− (q) Enantiomeric
products
(C) CH3CH2–CH = O )Cº25(
OH →−
(r) H
DKK
α
α → < 1(Primary
isotopic effect)
(D) CH3–
3CH|
–CH
O||
H–C →−OH.Conc (s) Diastereomeric
products
(t) Disproportionation
MATHEMATICS
SECTION – I Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
XtraEdge for IIT-JEE APRIL 2010 56
1. Let F denote the set of all onto functions from A = a1, a2, a3, a4 to B = x, y, z. A function f is chosen at random from F. The probability that f –1 (x) consists of exactly two elements is
(A) 2/3 (B) 1/3 (C) 1/6 (D) 0
2. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters TA are visible . The probability that the letter has come from CALCUTTA is
(A) 4/11 (B) 1/3 (C) 5/12 (D) None
3. A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelops at random, the probability that not all letters are placed in correct envelopes is
(A) 1/24 (B) 11/24 (C) 5/8 (D) 23/24
4. The value of x for which the matrix
A =
− 121010702
is inverse of
B =
−−
−
x2x4x010x7x14x
is
(A) 21 (B)
31 (C)
41 (D)
51
5. The largest term in the expansion of (3 + 2x)50, where x = 1/5, is
(A) 5th (B) 6th (C) 8th (D) 9th
6. If z2z − = 2, then the greatest value of | z | is
(A) 1 + 2 (B) 2 + 2
(C) 3 +1 (D) 5 + 1
7. The value of ∫ −−16
1
1 dx1xtan is
(A) 3
16π + 32 (B) π34 – 32
(C) π34 + 32 (D) π
316 – 32
8. The solution of (y + x + 5)dy = (y – x + 1) dx is
(A) log ((y + 3)2 + (x + 2)2) + tan–1 2x3y
++ = C
(B) log ((y + 3)2 + (x – 2)2) + tan–1 2x3y
−− = C
(C) log ((y + 3)2 + (x + 2)2) + 2 tan–1 2x3y
++ = C
(D) log ((y + 3)2 + (x + 2)2) – 2 tan–1 2x3y
++ = C
SECTION – II
Multiple Correct Answers Type
This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 9. If n > 1, then (1 + x)n – nx – 1 is divisible by - (A) x2 (B) x3 (C) x4 (D) x5
10. If y = 1A4sin11A4sin1
−+
+− , then one of the values of y is -
(A) – tan A (B) cot A
(C) tan
+
π A4
(D) – cot
+
π A4
11. Equation of a common tangent to the circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0 is - (A) x = 1 (B) x = 0 (C) x + y3 + 3 = 0 (D) x – y3 + 3 = 0
12. The points of extremum of ∫ ++−2x
0 t
2
e24t5t are -
(A) x = – 2 (B) x = 1 (C) x = 0 (D) x = – 1
SECTION – III Comprehension Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.
Paragraph # 1 (Ques. 13 to 15) Consider the region S0 which is enclosed by the curve
y ≥ 2x1− and max. |x|, |y| ≤ 4. If slope of a family of lines is defined as m(t) = cos t where point (t, 2t + 0.4) lies inside the region S0. Any member of this family of lines is called L1= 0 if it passes through (π, max t) and L2 = 0 if it passes through the (π, minm t).
13. Area of region of S0 is - (A) 8 + π/2 sq. units (B) 8 – π/2 sq. units (C) 8 sq. units (D) 8 + π sq. units
XtraEdge for IIT-JEE APRIL 2010 57
14. If line lies inside the region S0, then - (A) t ∈ (0, 0.28) (B) t ∈ (0, 1) (C) t ∈ (0.28, 1) (D) t ∈ [.28, 1]
15. L1 = 0 having maximum slope, is -
(A) π−
−x
1y = cos 1 (B) π−
−x
1y = cos (0.28)
(C) 1x
y−
π− = cos (1) (D) None of these
Paragraph # 2 (Ques. 16 to 18) Two circles S1 = 0 and S2 = 0 are touching to each
other externally at point T, with centre C1, C2 and radii r1 and r2 respectively.
If P and Q be the points of contact of a direct common tangent to the two circles and PQ meets the line joining C1, C2 in S. Tangent at common point T is intersecting to the tangent PQ at R point and to other direct tangent at V point. Let S1= x2 + y2 – 6x = 0 and S2= x2 + y2 + 2x = 0.
16. Angle between the two direct tangents is– (A) 90º (B) 30º (C) 60º (D) None of these
17. Direct tangents are–
(A) y = 3 x + 3 , y = – 3 x + 3
(B) y = 3
x – 3 , y = 3x− – 3
(C) y = 3
x + 3 , y = 3x− – 3
(D) None of these 18. A circle S = 0 of radius 1 units rolls on the outside of
the circle S2 = 0, touching it externally, locus of the centre of this outer circle is –
(A) Circle (B) Ellipse (C) Parabola (D) None of these
SECTION – IV
Matrix – Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
19. Match the following-:
Column- I Column- II
(A) The sum of the series
∑=
10
0rr
20 C is
(p) – 10C5
(B) The coefficient of x53 in
∑=
−−100
0r
rr100r
100 2)3x(C is
(q) 100C4
(C) (10C0)2 –(10C1)2 +……. …..–(10C9)2 + (10C10)2 equals
(r) 219+21 20C10
(D) The value of
95C4 + ∑=
−5
1j3
j100 C is
(s) – 100C53
(t) 100C47
20. Match the following -
Column -I
Column -II
(A) If the lines 1
2x − = 1
3y − = λ− 4z
and λ−1x =
24y − =
15z −
intersect at (α, β, γ) then λ =
(p) 0
(B) If ∞→x
lim 4x
++
−π −
2x1xtan
41 =
y2 + 4y + 5 then y =
(q) –1
(C) If chord x + y + 1= 0 of parabola y2 = ax subtends 90º at (0, 0) then a =
(r) –2
(D) If a = i + j + k , a . b = 1
and a × b = j – k , then | b | is equal to
(s) 1
(t) –3
XtraEdge for IIT-JEE APRIL 2010 58
PHYSICS
SECTION – I Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1. ABCD is a smooth horizontal fixed plane on which mass m1 = 0.1 kg is moving in a circular path of radius r = 1 m. It is connected by an ideal string which is passing through a smooth hole and connects
of mass m2 = 1/ 2 kg at the other end as shown. m2 also moves in a horizontal circle of same radius of 1
m with a speed of 10 m/s. If g = 10 m/s2 then the speed of m1 is-
m2
D
m1
A B
C
(A) 10 m/s (B) 10 m/s
(C) 101 m/s (D) None of these
2. A L shaped rod whose one rod is horizontal and other is vertical is rotating about a vertical axis as shown with angular speed ω. The sleeve shown in figure has mass m and friction coefficient between rod and sleeve is µ. The minimum angular speed ω for which sleeve cannot sleep on rod is –
m
ω
sleeve
l
(A) lµ
=ωg (B)
l
gµ=ω
(C) gµ
=ωl (D) None of these
3. Two solid spherical balls of radius r1 & r2
(r2 < r1), of density σ are tied up with a string and released in a viscous liquid of lesser density ρ and coefficient of viscosity η, with the string just taut as shown. The terminal velocity of spheres is -
r1
r2
(A) )(gr92 2
2 ρ−ση
(B) )(gr92 2
1 ρ−ση
(C) ηρ−σ
++ g)(
rr)rr(
92
21
32
31 (D)
ηρ−σ
−− g)(
rr)rr(
92
21
32
31
4. A block of mass m is attached to an ideal spring and system lies in vertical plane as shown. Initially the supporting plane is placed so that spring remains in its natural length then the plane is moved very slowly downwards. The graph showing variation of normal reaction applied by mass on supporting plane with distance travelled by block is –
Supporting plane
M
(A)
mg
x
N
(B)
mg x
N
(C)
mg
x
N
(D) None of these
5. A massless container is filled with liquid of density ρ. It contains two holes as shown in figure. Container rests on ground. Area of the two holes are A each. Container is filled with liquid upto height H. Then –
H3H/4H/4
(A)Torque produced by normal force between container & ground about center of gravity is
2AgH2ρ into the plane of paper
(B) Torque produced by friction about center of gravity is ρAgH2 out of the plane of paper
(C) Net torque produce by thrust force & friction
force about center of gravity is 4
AgH2ρ into the
plane of paper (D) Torque produced by normal force between
container and ground about centre of gravity is zero
XtraEdge for IIT-JEE APRIL 2010 59
6.
A B
Two containers A & B contain ideal gases helium and oxygen respectively. Volume of both containers are equal and pressure is also equal. Container A has twice the number of molecules than container B then if vA & vB represent the rms speed of gases in containers A & B respectively, then -
(A) B
A
vv = 2 (B)
B
A
vv = 4
(C) B
A
vv = 2 (D)
B
A
vv = 8
7. A capacitor is composed of three parallel conducting plates. All three plates are of same area A. The first pair of plates are kept a distance d1 apart and the space between them is filled with a medium of a dielectric ε1. The corresponding data for the second pair are d2 & ε2 respectively. What is the surface charge density on the middle plate ?
d1 d2
ε1 ε2
V0
(A)
ε+
εε
2
2
1
10 ddV (B)
ε+
εε−
2
2
1
10 ddV
(C)
ε+
εε
2
2
1
10 ddV2 (D)
ε+
εε−
2
2
1
10 ddV2
8. The mirror of length 2l makes 10 revolutions per minute about the axis crossing its mid point O and perpendicular to the plane of the figure There is a light source in point A and an observer at point B of the circle of radius R drawn around centre O (∠AOB = 90º)
What is the proportion l
R if the observer B first sees
the light source when the angle of mirror ψ = 15º ?
A
R Bl
l
O
ψ
(A) 2 (B)2
1 (C) 22 (D) 22
1
SECTION – II
Multiple Correct Answers Type
This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
9. A body moves in a circular path of radius R with deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment t = 0, the velocity of body is v0 then the velocity of body at any time will be –
(A) v =
+
Rtv1
v
0
0 at time t
(B) v = RS
0ev−
after it has moved S meter
(C) v = v0e–SR after it has moved S meter (D) None of these
10. A cylinder block of length L = 1m is in two
immiscible liquids. Part of block inside liquid(1) is
41 m and in liquid (2) is
41 m. Area of cross-section
of block is A. Densities of liquid (1) & (2) are ρ and 2ρ respectively –
Liquid (1) ρ.
2ρ. Liquid (2)
XtraEdge for IIT-JEE APRIL 2010 60
(A) Density of block is 3ρ/4
(B) Force exerted by liquid (1) on block is ρAg/4 (C) Block is depressed so that it is just completely
immersed in liquid (1) and released. A initial acceleration of block is 4/3 g
(D) In case (C) force exerted by liquid (2) on block is 3/2 ρAg
11. R = 10Ω & E = 13 V and voltmeter & Ammeter are ideal then -
V
A
c 6V
b 3Ω
8V
a
R
E
(A) Reading of Ammeter is 2.4 A (B) Reading of Ammeter is 8.4 A (C) Reading of voltmeter is 8.4 V (D) Reading of voltmeter is 27 V 12. A parallel plate air capacitor is connected to a
battery. If plates of the capacitor are pulled further apart, then which of the following statements are correct -
(A) Strength of electric field inside the capacitor remain unchanged, if battery is disconnected before pulling the plate.
(B) During the process, work is done by an external force applied to pull the plates whether battery is disconnected or it remain connected.
(C) Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart.
(D) None of the above
SECTION – III Comprehension Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.
Paragraph # 1 (Ques. 13 to 15) In the shown arrangement, both springs are relaxed.
The coefficient of friction between m2 and m1 is µ. There is no friction between m1 and surface. If the blocks are displaced slightly they perform SHM together
m2 m1
k2 k1
13. If the small displacement of blocks is x then
acceleration of m2 is-
(A) 2
2
mxk
(B) 2
21
mx)kk( +
(C) xmmkk
21
21
++ (D) None of these
14. The condition in which frictional force on m2 acts in the direction of its displacement from mean position is –
(A) 2
1
2
1
kk
mm
> (B) 2
1
1
2
kk
mm
>
(C) 2
1
2
1
kk
mm
= (D) None of these
15. If the condition obtained in Q.15 is met, then the maximum amplitude of oscillation is –
(A) 1221
212
kmkm)mm(gm
−+µ (B)
2211
212
kmkm)mm(gm
−+µ
(C) 1221
212
kmkm)mm(gm
++µ (D) None of these
Paragraph # 2 (Ques. 16 to 18) A conducting rod PQ of mass M rotates without
friction on a horizontal plane about Ο on circular rails of diameter 'l'. The centre O and the periphery are connected by resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At t = 0, PQ starts rotating clockwise with angular velocity ω0. Neglect the resistance of the rails and rod, as well as self inductance.
XtraEdge for IIT-JEE APRIL 2010 61
B
O ω0
Q
R P
⊗
16. Magnitude of current as a function of time
(A) t2
0 eR2
B α−ω l (B) t22
0 eR16
B α−ω l
(C) t2
0 eR8
B α−ω l (D) t22
0 eR8
B α−ω l
Where α = RM8B3 22l
17. Total charge flow through resistance till rod PQ stop rotating .
(A) B8M0ω (B)
B3M0ω (C)
B6M0ω (D)
B9M0ω
18. Heat generated in the circuit by t = ∞
(A) 24
M 20
2ωl (B) 8
M 20
2ωl
(C) 3
M 20
2ωl (D) 32
M 20
2ωl
SECTION – IV Matrix – Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
19. A uniform solid cube is floating in a liquid as shown in the figure, with part x inside the liquid. Some changes in parameters are mentioned in Column I. Assuming no other changes, match the following -
x
Column I Column II (A) If density of the liquid (p) Increase decreases, x will (B) If height of the cube is (q) Decreases increased keeping base area and density same, x will (C) If the whole system is (r) Remain same accelerated upward, then x will (D) If the cube is replaced (s) May increase by another cube of same or decrease size but lesser density, x will (t) none 20. A block of mass m = 1 kg is at rest with respect to a
rough wedge as shown in figure.
µm
a
θ The wedge starts moving up from rest with an
acceleration of a = 2m/s2 and the block remains at res with respect to wedge then in 4 sec. of motion of wedge work done on block (assume angle of inclination of wedge is θ = 30º and g = 10 m/s2) –
Column I Column II
(A) By gravity (p) 144 J (in magnitude)
(B) By normal reaction (q) 32 J
(C) By friction (r) 160 J
(D) By all the forces (s) 48 J
(t) none
XtraEdge for IIT-JEE APRIL 2010 62
CHEMISTRY
SECTION – I Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1.
NH2
NOBr Product
The main product is –
(A)
Br
+ Enantiomer
(B)
Br
+ Enantiomer
(C) Br
+ Enantiomer
(D) NO
+ Enantiomer
2. There are two isomeric carboxylic acids– 'A' and 'B'
C9H8O2. reacts with H2/Pd giving compounds, C9H10O2. 'A' gives a resolvable product and 'B' gives a non-resolvable product. Both isomers could by oxidised to PhCOOH.
The correct structures of 'A' and 'B' are, respectively–
(A) CH2=CH COOH ;
CH=CHCOOH
(B) COOHCH2=C ; CH=CH–COOH
(C) CH=CHCOOH ; CH2=CH COOH
(D) CH=CH–COOH ;
COOHC=CH2
3. If H-He undergoes dissociation, which of the follwoing product most expected to occur –
2HHe products (A) He2 + 2H (B) H2 + 2He (C) D2 + He2 (D) 2H + 2He
4. 100 ml solution (I) of buffer containing 0.1(M) HA and 0.2 (M) A–, is mixed with another solution (II) of 100 ml containing 0.2(M) HA and 0.3(M) A– .After mixing what is the pH of resulting solution ?( Given pKa of HA = 5)
(A) 5 – log 5/3 (B) 5 + log 5/3 (C) 5 + log 2/5 (D) 5 – log 5/2
SECTION – II Multiple Correct Answers Type
This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
MOCK TEST FOR IIT-JEE
PAPER - II
Time : 3 Hours Total Marks : 240
Instructions : • This question paper contains 57 questions in Chemistry (19,) Mathematics (19) & Physics (19). • In section -I (4 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (5 Ques) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer • In section -III (2 Ques.) of each paper +8(2×4) marks will be given for correct answer. No Negative marking for
wrong answer. • In section -IV (8 Ques.) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer.
XtraEdge for IIT-JEE APRIL 2010 63
5. Dopamine of a drug used in the treatment of parkinson's disease.
CH2–CH
Dopamin
NH2 COOH HO
HO
Which of the following statements about this compound are correct ?
(A) It can exist in optically active forms. (B) One mole will react with three moles of sodium
hydroxide to form a salt (C) It can exist as a Zwitter ion in the aqueous
solution (D) It gives nitroso compound on treatment with
HNO2.
6. Which of the following method(s) would be useful for preparing ketones ?
(A) Friedel-Crafts reaction of an acyl chloride with benzene (AlCl3 catalyst)
(B) Reaction of methyllithium with the lithium salt of carboxylic acid, followed by hydrolysis
(C) Reaction of R2CuLi with an acyl choride in ether at low temperature
(D) Reaction of Grignard reagents with acyl chloride in ether followed by hydroylsis
7. In the given table types of H bonds and some H bond energies are given and other H bond energies are not given. You are to perdit the unknown H-bond energies.
Types of H-bonds H-bond energies in KJ/mol (I) F – H …….. O – F – H …….. F 30 (II) O – H …….. O – O – H …….. F 15 (III) F – H …….. F– – (IV) N – H …….. N – Correct prediction are – (A) H-bond energy for (I) may be 20 kJ/mol (B) H-bond energy for (II) may be 25 kJ/mol (C) H-bond energy for (III) may be 113 kJ/mol (D) H-bond energy for (IV) may be 12 kJ/mol
8. During conductance measurement of an electrolyte based on the wheatstone bridge principle alternating current is used because direct current produces –
(A) polymerisation (B) Ionisation (C) Electrolysis resulting in the chnge of
concentration and in consequencae the resistance (D) Polarisation at the electrodes resulting in the
change of resistance
9. Consider the cell : Ag | AgCl (s) | KCl (1M) | Hg2Cl2(s) | Hg (l) | Pt. The
cell potential : (A) increases on increasing concentration of Cl– ions (B) decreases on decreasing concentration of Cl– ions (C) is independent of concentration of Cl– ions
(D) is independent of amounts of AgCl (s) and Hg2Cl2 (s)
SECTION – III
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
10. Match the following Column I Column II (A) HCOOH (p) Decarboxylation on heating (B) CH3COOH (q) Reation with Br2 (C) OH
COOH (r) Cu2+(alkaline)→Cu2O
(D) PhCH2COOH (s) Decarbonylation or decarboxylation on treatment with conc. H2SO4 (t) Reaction with I2+NaOH
11. Match the following Column I Column II (A) XeO4 (p) Non polar (B) XeF4 (q) Having no lone pair of electron in the central atom (C) SO3 (r) Planar (D) BF3 (s) Having lone pair of electrons in the central atom (t) Tetrahedral
XtraEdge for IIT-JEE APRIL 2010 64
SECTION – IV Integer answer type
This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
X Y Z W
12. 3 ampere current was passed through an aqueous
solution of an unknown salt of Pd for one hour, 2.977g of Pdn+ was deposited at cathode. Find n. (At. wt. of Pd = 106.4)
13. Consider the following reaction sequence
O glycol/KOH
NHNH 22 → W →OH/O 23 X
∆ → 2)OH(Ca
Y →COOOHHC 56 Z
How many carbons are present in the final product Z? 14. For the reaction in the plant cells 6CO2(g) + 12H2O (l) → C6H12O6(s) + 6O2(g) + 6H2O(l) ∆rGº = 3000 kJ/mol ATP → ADP + PO4
3– ∆rGº = – 30 kJ/mol Glucose is stored in the plant cell as starch,
(C6H10O5)n. To produce 162 g of starch how many moles of ATP are minimum required ? Give your answer after divide actual answer by 100.
15. The relative lowering of the vapour pressure of an
aqueous solution containing a non-volatile solute is 0.0125. What is the molality of the solution. Give your answer after multiplying actual answer by 10.
16. A polyvalent metal weighing 0.1g and having atomic
weight 51 reacted with dil. H2SO4 to give 43.9 ml of hydrogen at STP. The solution containing the metal in this lower oxidation state(X), was found to require 58.8 mL of 0.1 N KMnO4 for complete oxidation. What is the higher oxidation state (Y) of the metal ?
17. The stopcock, connecting the two bulbs of volumes 5 litres and 10 litres containing an ideal gas at 9 atm and 6 atm respectively, is opened. What is the final pressure (in atm) in the two bulbs if the temperature remained the same ?
18. Sum of lone pairs present in XeOF4 and XeO3 is ….. . 19. Total number of isomers are possible in [Co(en)2Cl2]
and [Co(C2O4)2(NH3)2]– is.
MATHEMATICS
SECTION – I Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
1. Let ∑=
=n
1r
4 )n(fr , then ∑=
−n
1r
4)1r2( is equal to
(A) f(2n) – 16 f(n) (B) f (2n) – 7f (n) (C) f(2n – 1) – 8 f(n) (D) None of these
2. If fr(α) =
α
+α
22 rsini
rcos ×
α
+α
22 r2sini
r2cos ….
α
+α
rsini
rcos
then nnflim
∞→(π) equals
(A) –1 (B) 1 (C) – i (D) i
3. If α, β, γ, δ are four complex numbers such that δγ is
real and αδ – βγ ≠ 0, then z = tt
δ+γβ+α ,
t ∈ R represents a (A) circle (B) parabola (C) ellipse (D) straight line 4. The inflection points on the graph of function
y = ∫ −−x
0
2 dt)2t)(1t( are
(A) x = –1 (B) x = 3/2 (C) x = 4/3 (D) x =1
SECTION – II Multiple Correct Answers Type
This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
XtraEdge for IIT-JEE APRIL 2010 65
5. If A and B are two invertible matrices of the same order, then adj (AB) is equal to -
(A) adj (B) adj (A) (B) |B| |A| B–1A–1 (C) |B| |A| A–1B–1 (D) |A| |B| (AB)–1
6. If A (α , β) =
αα−αα
βe000cossin0sincos
, then
(A) A (α, β)′ = A (– α, β) (B) A (α, β)–1 = A (–α, –β) (C) Adj (A (α, β)) = e–β A (–α, –β) (D) A (α, β)′ = A (α, – β) 7. For a positive integer n, if the expansion of
n4
2 xx5
+ has a term independent of x, then n can
be (A) 18 (B) 21 (C) 27 (D) 99 8. If k is odd then kCr is maximum for r equal to
(A) 21 (k – 1) (B)
21 (k + 1)
(C)k – 1 (D) k 9. Let an =
43421timesn
)1...111( , then
(A) a912 is not prime (B) a951 is not prime (C) a480 is not prime (D) a91 is not prime
SECTION – III Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
10. Match the following Column-1 Column-II Equation General Solutions
(A) 2 sin θ – 3 = 0 (p) nπ + (–1)n 3π
(B) 2 sin 2θ + 3 (q) 2nπ –3π
= 2 sin θ + 2 3 cos θ
(C) sin 2θ + cos 2θ + 4 sin θ (r) 2nπ +3π
= 1 + 4 cos θ
(D) cos2θ = 41 (s) nπ –
4π
(t)
nπ +
4π
11. Match the following :
Column- I Column- II
(A) If three unequal numbers a, b, c are in A.P. and b – c, c – b, a are in G.P., then
abc3cba 333 ++ is equal to
(p) 1/3
(B) Let x be the arithmetic mean and y,z be two geometric means between any two positive numbers, then
xyzzy 33 + is equal to
(q) 1
(C) If a, b, c be three positive number which form three successive terms of a G.P. and c> 4b –3a, then the common ratio of the G.P. can be equal to
(r) 2
(D) ∞→n
lim tan
∑
=
−n
1r2
1
r21tan
is equal to
(s) 3
(t) 1/2
SECTION – IV Integer answer type
This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :
XtraEdge for IIT-JEE APRIL 2010 66
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
X Y Z W
12. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the value of 2 (z + z ) – |z|2.
13. Find the sum of all the integral roots of (log5 x)2 + log5x (5/x) = 1.
14. Four different integers form an increasing A.P. such that one of them is the square of the remaining numbers. Find the largest numbers.
15. If n is a positive integer, and E = 2n + 1C1 + 2n + 1C2 + … + 2n + 1Cn – 2n + 1C2n + 1 – 2n + 1C2n – … – 2n + 1Cn + 1. Find |E|
16. Find the number of values of t for which the system of equations
(a + 2t)x + by + by + cz = 0 bx + (c + 2t)y + az = 0 cx + ay + (b + 2t) z = 0 has non-trivial solutions.
17. A function f(x) is defined for x > 0 and satisfies f(x2) = x3 for all x > 0. Then the value of f ′(4) is ___.
18. If A is the area formed by the positive x-axis, and the normal and tangent to the circle x2 + y2 = 4 at (1, 3 ) then A/ 3 is equal to _____.
19. The area bounded by the curves x = y2 and x = 3 – 2y2 is ______.
PHYSICS
SECTION – I Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A thin conducting plate is inserted in half way between
the plates of a parallel plates capacitor of capacitance C.
Conducting plate
What does the value of capacitance, if both the plate of capacitor is shortened ?
(A) C (B) 2C (C) 3C (D) 4C 2. White light is incident normally on a glass surface
(n = 1.52) that is coated with a film of mg F2 (n = 1.38). For what minimum thickness of the film will yellow light of wavelength 550 nm (in air) be missing in the reflected light ?
mgF2 glass
(A) 99.6 nm (B) 49.8 nm (C) 19.6 nm (D) 10.6 nm 3. A circuit element is placed in a closed box. At time
t = 0, a constant current generator supplying a current of I amp is connected across the box. Potential diff. across the box varies according to graph as shown in the figure. The element in the box is -
3 2
8Volts(V)
Time t(sec)
(A) a resistance of 2Ω (B) a battery of emf 6V (C) an inductance of 2H (D) a capacitance 4. Consider a usual set-up of Young's double slit
experiment with slits of equal intensity as shown in the figure. Take 'O' as origin and the Y axis as indicated. If average intensity between y1 = λD/4d and y2 = λD/4d equals n times the intensity of maximum, then n equal is (take average over phase difference) -
S2
d
S1
O
y
D
(A)
π+
2121 (B) 2
π+
21
(C)
π+
21 (D)
π−
2121
SECTION – II
Multiple Correct Answers Type
This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.
XtraEdge for IIT-JEE APRIL 2010 67
5. A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as ρ
= 2rα , α is a constant. The total resistance per unit
length of the conductor is R and the electric field strength in the conductor due to which a current I flows in it is E.
(A) R = 2A2πα
(B) R = 2A4πα
(C) E = 2AI2πα (D) E = 2A
I4πα
6. Two infinite plates carry j ampere of current out of the age per unit width of the plate as shown. BP and BQ represent magnitude of field at points P and Q respectively.
P
Q
(A) BP = 0 (B) BP = µ0j (C) BQ = 0 (D) BQ = µ0j
7. A bar magnet M is allowed to fall towards a fixed conducting ring C. If g is the acceleration due to gravity, v is the velocity of the magnet at t = 2 s and s is the distance traveled by it in the same time then,
M
C
3g
(A) v > 2g (B) v < 2g (C) s > 2g (D) s < 2g 8. In the network shown, the capacitor C is initially
uncharged. The time constant of the circuit is τ and the charge on C at time t after the switch S is closed is q.
C S
R3
R2
R1
(A) τ = CR1 (B) τ = C
+
+32
321 RR
RRR
(C) q = )e1(RR
CVR /t
32
2 τ−−+
(D) q = )e1(RR
CVR /t
21
1 τ−−+
9. Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is -
0
Bin
ding
Ene
rgy/
N
ucle
on in
MeV
5.0
7.58.08.5
30 60 90 120Mass number of nuclei
Z
Y X
W
(A) Y → 2Z (B) W → X + Z (C) W → 2Y (D) X → Y + Z
SECTION – III Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.
q r
p s
r
p q
t
s t
r
A
B
C
D
p q r s t
10. Capillary rise and shape of droplets on a plate due to surface tension are shown in column II.
Column I Column II
(A) Adhesive forces is (p)
B
A
greater than cohesive forces
(B) Cohesive forces is (q)
A
B
greater than adhesive forces
XtraEdge for IIT-JEE APRIL 2010 68
(C) Pressure at A > pressure (r) A mercury drop at B is pressed between two parallel plates of glass
AB
(D) Pressure at B > Pressure (s) AB
at A (t) none 11. Column I Column II (A) The coefficient of (p ) with decrease volume expansion at in pressure constant pressure is (B) Mean free path of (q) at all temperatures molecule increases (C) An ideal gas obeys (r) Same for all gases Boyle’s and Charle’s Law (D) A real gas behaves (s) At high temperature as an ideal gas at low pressure and (t) none
SECTION – IV Integer answer type
This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
0 1 2
3
4
5 6
7
8
9
X Y Z W
12. A ball A is falling vertically downwards with
velocity v1. It strikes elastically with a wedge moving horizontally with velocity v2 as shown in figure. Find
the value of 2
1
vv , so that the ball bounces back in
vertically upward direction relative to the wedge.
v2
A
v1
30º
13. Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC, then find the value of ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches to point C.
A
D
B
C
14. In what minimum time after its motion begins will a
particle oscillating according to the law x = 7sin21
πt
move from the position of equilibrium to the position x = – 7/2 units ?
15. A satellite is revolving in a circular equatorial orbit of radius R = 2 × 104 km from east to west. Calculate the interval (in hrs) after which it will appear at the same equatorial town. Given that the radius of the earth = 6400 km and g (acceleration due to gravity) = 10 ms–2.
16. Three capillary tubes of same radius 1 cm but of lengths 1m, 2m and 3m are fitted horizontally to the bottom of a long cylinder containing a liquid at constant pressure and flowing through these tubes. What is the length of a single tube which can replace the three capillaries.
17. Equations of a stationary and a traveling waves are as follows : y1 = a sin kx cos ωt and y2 = a sin (ωt – kx)
The phase difference between two points x1 = k3π and
x2 = k23π are φ1 and φ2 respectively for the two waves.
Then find the value of 2
1
φφ .
18. A coil, a capacitor and an AC source of voltage 24 V (rms) are connected in series. By varying the frequency of the source, a maximum rms current of 6A is observed. If this coil is connected to a battery of emf 12 V and internal resistance 4 Ω, then find the value of current flow through it.
19. A potential difference of 103 V is applied across an
X-ray tube. Calculate the value of ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced.
XtraEdge for IIT-JEE APRIL 2010 69
PHYSICS
1. An P-N-P transistor circuit is arranged as shown. It is a –
RL = 10 K V
V
PN P
(A) common base amplifier circuit (B) common-emitter amplifier circuit (C) common-collector circuit
(D) None 2. A tuning fork and an air column in resonance tube
whose temperature is 51°C produces 4 beats in 1 second when sounded together. When the temperature of the air column decreases, the number of beats per second decreases. When the temperature remains 16°C, only 1 beat per second is produced. Then the frequency of the tuning fork is -
(A) 55 Hz (B) 50 Hz
(C) 68 Hz (D) none 3. For wave propagation wrong statement is - (A) The wave intensity remains constant for a plane
wave
(B) The wave intensity decreases as the inverse of the distance from the source for a spherical wave
(C) The wave intensity decreases as the inverse square of the distance from the source for spherical wave
(D) Total intensity of the spherical wave over the spherical surface centered at the source remains constant at all times
4. Statement-1: Temperature of the body is lowered considerably if we put wet clothes.
Statement -2: Specific heat of water is high. (A) Both statement -1 and statement -2 are true and statement -2 is a correct explanation of the statement -1
(B) Both statement -1 and statement -2 are true but statement -2 is not a correct explanation of the statement -1 (C) Both statement -1 and statement -2 are false (D) statement -1 is false but the statement -2 is true.
5. An observer starts moving with uniform acceleration a toward a stationary sound source emitting a whistle of frequency n. As the observer approaches source, the apparent frequency n' heard by the observer varies with time as –
(A)
time
n'
(B)
time
n'
(C)
time
n'
(D)
time
n'
6. The main scale of a spectrometer is divided into 720 divisions in all. If the vernier scale consists of 30 divisions, the least count of the instrument is (30 division of vernier scale coincide with 29 division of main scale) -
(A) 0.1° (B) 1'' (C) 1' (D) 0.1''
SYLLABUS : Physics : Full syllabus Chemistry : Full syllabus Mathematics : Full syllabus
Time : 3 Hours Total Marks : 432
Instructions : • Part A – Physics (144 Marks) – Questions No. 1 to 2 and 9 to 30 consist FOUR (4) marks each and Question No.
3 to 8 consist EIGHT (8) marks each for each correct response. Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 46 to 60 consist FOUR (4) marks each and
Question No. 40 to 45 consist EIGHT (8) marks each for each correct response. Part C – Mathematics (144 Marks) – Questions No.61 to 82 and 89 to 90 consist FOUR (4) marks each and
Question No. 83 to 88 consist EIGHT (8) marks each for each correct response • For each incorrect response, ¼ (one fourth) of the weightage marks allotted of the would be deducted.
MOCK TEST - AIEEE PATTERN
XtraEdge for IIT-JEE APRIL 2010 70
7. Experimental verification of Newton's law of cooling is valid for -
(A) large temperature difference i.e. 30°C to 85°C between hot liquid and surrounding
(B) very large temperature difference i.e. 5°C to 95°C between hot liquid and surrounding
(C) small temperature difference i.e. 30°C to 35°C between hot liquid and surrounding
(D) any temperature difference
8. While studying the dissipation of energy of a simple pendulum by plotting a graph between square of amplitude and time which of the following apparatus is not essential ?
(A) Ticker timer (B) Meter scale (C) Vernier calliper (D) Stop watch
9. An object is weighed on a balance whose pans are not equal in masses when placed in the left pan, the object appears to weigh 10.30g but when placed in the right pan, it appears to weigh 12.62g. What is the correct mass of the object ?
(A) 10.30 g (B) 12.62 g (C) 11.46 g (D) Can not find
10. When jockey is put at two ends of the potentiometer wire, the galvanometer gives diflections in opposite directions. It means that apparatus can -
(A) not give a null point
(B) give a null point (C) be faulty
(D) be used after making some changes in the circuit
11. A student performs an experiment to determine the Young's modulus of a wire exactly 2cm long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is -
(A) (2.0 ± 0.3) × 1011 N/m2
(B) (2.0 ± 0.2) × 1011 N/m2 (C) (2.0 ± 0.1) × 1011 N/m2
(D) (2.0 ± 0.05) × 1011 N/m2
12. A student measures the focal length of a convex lens by putting an object pin at a distance 'u' from the lens and measuring the distance 'v' of the image pin. The graph between 'u' and 'v' plotted by the student should look like -
(A)
v(cm)
O u(cm)
(B) O
v(cm)
u(cm)
(C)
v(cm)
O u(cm)
(D)
v(cm)
Ou(cm)
13. Two concave mirror each of focal length f. A point source is placed at a point midway between two mirror. The minimum value of d for which only one image of s is formed –
d
s
(A) f (B) 2f
(C) 3f (D) 4f 14. In YDSE, if the intensity of central maxima is IO then
the y-coordinate of point where the intensity is 2
IO ,
d = 0.1 mm, D = 1m, λ = 5000A°
d
P
y
O
D
(A) 1.5 mm (B) 2 mm (C) 1.75 mm (D) 1.25 mm 15. Two charges of –4µC and +4µC are placed at points
A (1,0,4) and B (2, –1,5) located in an electric field →E = 0.20 i V/cm. The torque acting on the dipole
is-
(A) 8 × 10–5 N-m (B) 8/ 2 × 10–5 N-m
(C) 8 2 × 10–5 N-m (D) 22 × 10–5 N-m 16. Three concentric spherical metallic shells A, B and C
of radii a, b and c (a < b < c) have surface charge densities σ, – σ and σ respectively. If the shells A and C are at same potential, then the correct relation between a, b and c is -
(A) a + c = b (B) b + c = a (C) a – b = c (D) a + b = c
XtraEdge for IIT-JEE APRIL 2010 71
17. Five identical plates of equal area A are placed parallel to and at equal distance d from each other as shown in figure. The effective capacity of the system between the terminals A and B is -
B
A
(A) 53
dAo∈
(B) 45
dAo∈
(C) 35
dAo∈
(D) 54
dAo∈
18. Read the following statements carefully : Y : The resistivity of semiconductor decreases with
increase of temperature. Z : In a conducting solid, the rate of collision
between free electrons and ions increases with increase of temperature.
(A) Both Y and Z are true and Z is correct explaination of Y
(B) Both Y and Z are true but Z is not correct explaination of Y
(C) Y is true but Z is false
(D) Y is false but Z is true
19. A 1m long metallic wire is broken into two unequal parts A and B. The part A is uniformly extended into another wire C. The length of C is twice the length of A and resistance of C is equal to that of B. The ratio of resistances of parts A and C is -
(A) 4 (B) 41
(C) 2 (D) 21
20. A 600 cm long potentiometer wire is connected to a circuit as shown in figure. The resistance of potentiometer wire is 15r. The distance from point A at which the jockey should touch the wire to get zero deflection in the galvanometer is –
G
E
A B E/2 r
r
J
(A) 320 cm (B) 230 cm (C) 160 cm (D) 460 cm 21. A rectangular loop of metallic wire is of length a and
breadth b and carries a current i. The magnetic field at the centre of the loop is -
(A) π
µ4
i0 ab
ba8 22 + (B)
πµ4
i0ab
ba4 22 +
(C) π
µ4
i0ab
ba2 22 + (D) π
µ4
i0ab
ba 22 +
22. A short magnet produces a deflection of 30° when placed at certain distance in tanA position of magnetometer. If another short magnet of double the length and thrice the pole strength is placed at the same distance in tanB position of the magnetometer, the deflection produced will be -
(A) 60° (B) 30°
(C) 45° (D) None of these
23. A solenoid has 2000 turns wound over a length of 0.30 m. Its area of cross-section is 1.2 × 10–3 m2
. Around its central section a coil of 300 turns is wound. If an initial current of 2A in the solenoid is reversed in 0.25 sec, the emf induced in the coil is equal to -
(A) 6 × 10–4 Volt (B) 4.8 × 10–2 Volt (C) 6 × 10–2 Volt (D) 48 kV
24. A 100 volt AC source of frequency 500 Hz is connected to a L–C-R circuit with L = 8.1 mH, C = 12.5 µF and R = 10 Ω, all connected in series. The potential difference across the resistance is -
(A) 100 V (B) 200 V
(C) 300 V (D) 400 V
25. Which one is correct ? (A) Resultant of two vectors of unequal magnitude
can be zero
(B) Resultant of three non-coplanar vectors of equal magnitude can be zero
(C) Resultant of three coplanar vectors is always zero (D) Minimum number of non-coplanar vectors whose
resultant can be zero is four.
26. A stone thrown with the velocity V0 = 14 m/s at an angle 45° to the horizontal, dropped to the ground at a distance 'S' from the point where it was thrown. From what height should the stone be thrown in horizontal direction with the same initial velocity so that it fall at the same spot -
(A) 14.2 m (B) 16.9 m (C) 10.0 m (D) 9.6 m
27. A small body of mass 'm' is attached to one end of a light inelastic string of length l. The other end of the string is fixed. The string is held initially taut and horizontal and then body is released. The centripetal acceleration of the body and the tension in the string when the string reaches vertical position will be -
(A) g, mg (B) 2g, 3 mg (C) 3g, 2mg (D) 3g, 3 mg
XtraEdge for IIT-JEE APRIL 2010 72
28. Assertion : A rocket moves forward by pushing the surrounding air backwards.
Reason : It derives the necessary thrust to move forward, according to Newton's third law of motion.
(A) Both Assertion and Reason are true and Reason is a correct explanation of the Assertion
(B) Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion
(C) Both Assertion and Reason are false
(D) Assertion is false but the Reason is true
29. While slipping on rough spherical surface of radius
'R', block A of mass 'm' comes with velocity gR4.1 at bottom B. Work done in slipping the block from 'B' to 'C' is –
m
A
B
C
(A) 4
mgR (B) mgR
(C) 1.3 mgR (D) 45 mgR
30. A 2000 kg rocket in free space expels 0.5 kg of gas per second at exhaust velocity 400 ms–1 for 5 seconds. What is the increase in speed of rocket in this time -
(A) 2000 ms–1 (B) 200 ms–1
(C) 0.5 ms–1 (D) zero
CHEMISTRY
31. Which of the following can act as a both Bronsted acid & Bronsted base -
(A) Na2CO3 (B) OH– (C) HCO3
– (D) NH3 32. In which compound the oxidation No. of Oxygen
is +21 -
(A) OF2 (B)O2F2 (C) O2[PtF6] (D) KO2 33. The favourable conditions for a spontaneous
reactions are - (A) T ∆S > ∆H, ∆H = ⊕ , ∆S = ⊕ (B) T ∆S > ∆H, ∆H = ⊕ , ∆S = Θ (C) T ∆S = ∆H, ∆H = Θ , ∆S = Θ (D) T ∆S = ∆H, ∆H = ⊕ , ∆S = ⊕
34. The substance not likely to contains CaCO3 is - (A) Dolomite (B) A marble statue (C) Calcined Gypsum (D) Sea shells
35. On mixing 10ml of acetone with 50ml of CHCl3, the total volume of the solutions-
(A) < 60ml (B) > 60ml (C) = 60ml (D) unpredictable
36. On addition of He gas at constant volume to the reaction N2 + 3H2 2NH3 at equilibrium-
(A) The reaction stops (B) Forward reactions is favoured (C) Reaction remains unaffected (D) Backward reactions is favoured
37. The half life of a reaction is 24 hours . If we start with 10gm of reactant, How many grams of it will reaction after 96 hours, (I order reaction)
(A) 0.625gm (B) 6.25gm (C) 1.25gm (D) 0.125gm
38. The current is passed in Ag2SO4 aqueous solution & 1.6 gm O2 is obtained. The amount of Ag deposited will be- [Ag = 108gm]
(A) 107.8g (B) 1.6g (C) 0.8g (D) 21.6g
39. In decinormal solution CH3COOH is ionized to the extent of 1.3% find the pH of solution.
(A) 3.89 (B) 2.89 (C) 4.89 (D) 5.89 40. A FCC element (atomic wt.= 60) has a cell edge of
400pm. Its density is- (A) 6.23 g/cm3 (B) 6.43 g/cm3 (C) 6.53 g/cm3 (D) 6.63 g/cm3 41. Which set of quantum No. is not possible- n l m s (A) 2 0 0 +1/2 (B) 4 2 –3 –1/2 (C) 3 2 –2 +1/2 (D) 2 1 0 +1/2 42. Give simplest formula of compound which
containing 6gm C, 3.01×1023 atom O and 2 mole H atoms-
(A) CH2O (B) CH4O (C) CHO (D) CH3O
43. The IUPAC Name of
(A) 1,2-dimethyl Cyclohexene (B) 2,3-dimethyl Cyclohexene (C) 1,2-dimethyl Cyclohex-2-ene (D) 5,6-dimethyl Cyclohex-1-ene 44. Which of the following reagent can make distinction
between Pri. and Sec. amines ? (A) NH3 (B) NaNO2/HCl (C) HCl (D) All
XtraEdge for IIT-JEE APRIL 2010 73
45. Toluene reacts with Cl2 in the presence of light to give -
(A) Benzyl chloride (B) Benzoyl chloride (C) p-chlorotoluene (D) o- chlorotoluene 46. Which compound is formed when excess of KCN is
added to an aqueous solution of copper sulphate (A) Cu (CN)2 (B) K2 [Ca(CN)4] (C) K [Cu(CN)2] (D) K3 [Cu(CN)4] 47. The Blue Print Process involves the use of- (A) Indigo dyes (B) Iron compound (C) Vat dyes (D) some other compounds 48. The ionic radii of N3–, O2–, F– and Na+ follow the
order- (A) N3– > O2– > F– > Na+ (B) N3– > Na+ > O2– > F– (C) Na+ > O2– > N3– > F– (D) O2– > F– > Na+ > N3– 49. Reaction xy2 xy + y (g) (g) (g) Initial pressure of xy2 is 600 mm Hg & total pressure
at equilibrium is 800 mm Hg. Kp of reaction is - (A) 50 (B) 100 (C) 166.6 (D) 400 50. Cell: Zn|Zn+2|| Cu+2| Cu If the correct reactions of Zn+2 & Cu+2 ions are
doubled, the emf of the cells: (A) doubled (B) halved (C) same (D) zero 51. What is the pH of buffer solution containing 12g
CH3COOH & 16.4g CH3COONa in 500ml of solution (Ka for CH3COOH = 1.8×10–5).
(A) 4.7447 (B) 4.4774 (C) 4.4477 (D) None 52. How many moles of ferrous oxalate are completely
oxidized by 1 mole KMnO4 in acidic medium- (A) 3/5 (B) 5/3 (C) 1/5 (D) 5 53. In an irreversible process taking place at constant-T
& P and which only pressure-volume work is being done, than (dG) and (dS), satisfy the criteria-
(A) (dS)V, E > 0, (dG)T,P < 0 (B) (dS)V, E = 0, (dG)T,P = 0 (C) (dS)V, E = 0, (dG)T,P > 0 (D) (dS)V, E < 0, (dG)T,P < 0 54. 3,3-dimethylbutan-2-ol, on reaction with Conc.
H2SO4 at 443K will give…… as major product- (A) 3,3-dimethyl but-1-ene (B) 2,3-dimethyl but-2-ene (C) 2,2-dimethyl but-2-ene (D) 2,2-dimethyl-1- butene
55. Select the true statement about benzene from amongst the following-
(A) Because unsaturation benzene easily undergoes addition reaction
(B) There are two types of C-C bonds in benzene molecule
(C) There is a cyclic delocalisation of π es– in benzene
(D) Monosubstitution of benzene group gives three isomeric substances
56. OH → dustZn B3
3
AlCl
ClCH → − K
→ 4KMnO.alk D. Identity ‘D’ ?
(A) CH3 (B)
(C) CHO
(D) COOH
57. In Lassaigne’s test, the organic compound is first fused with sodium metal. The sodium metal is used because (A) The melting point of sodium metal is low (B) Sodium metal reacts with elements present
in organic compounds to form inorganic compounds
(C) All sodium salts are soluble in water (D) All the above 58. Concentrated hydrochloric acid when kept in open air
sometimes produces a cloud at white fumes the explanation for it is that-
(A) Oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
(B) Strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke
(C) Due to strong affinity for water concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud.
(D) Concentrated hydrochloric acid emits strongly smelling HCl gas all the time
59. Consider the following complex [Co(NH3)5CO3]ClO4 The coordination number, oxidation number, number
of d-electrons and number of unpaired electrons on the metal respectively-
(A) 6, 3, 6, 0 (B) 7, 2, 7, 1 (C) 7, 1, 6, 4 (D) 6, 2, 7, 3
XtraEdge for IIT-JEE APRIL 2010 74
60. Point out the incorrect statement about resonance (A) Resonance structure should have equal energy (B) In Resonance structure, the constituent atoms
should be in the same position (C) In Resonance structure there should not be same
number of electron pairs (D) Resonance structure should differ only in the
location of electrons around the constituent atoms
MATHEMATICS
61. If the angles of elevation of an aeroplane from two points 1 km apart be 60º and 30º, then the height of the aeroplane is -
(A) 500m (B) 3
500 m
(C) 3
2000 m (D) None
62. Suppose a population A has 100 observations 101, 102, ..........200 and another population B has 100 observations 151, 152, ......... 250. If VA and VB
represents the variances of the two population
respectively then B
A
VV is -
(A) 49 (B)
94
(C) 32 (D) 1
63. If vertices of a triangle are (1, 0), (2, b) & (c2, – 3)
then centroid of triangle (A) can lie on y axis (B) always lie on x axis (C) lie on x axis if a + b = 3
(D) lie on y axis if c = 3 only
64. The value of k in order that f(x) = sin x – cos x – kx + b decreases for all real
values is given by : (A) k < 1 (B) k ≥ 1
(C) k > 2 (D) k < 2
65. 0xlim→ 2
x
xxcos–e
2
is equal to -
(A) 3/2 (B) 1/2 (C) 2/3 (D) None
66. If ∆ =
333
222
111
cbacbacba
, then
33333
22222
11111
cbc3b2acbc3b2acbc3b2a
−+−+−+
is equal to -
(A) ∆ (B) 2∆ (C) – 3∆ (D) 0 67. In a ∆ABC, angle A is greater than angle B. If the
measures of angles A and B satisfy the equation 3sin
x – 4 sin3x – k = 0, 0 < k < 1, then the measure of angle C is -
(A) 3π (B)
2π (C)
32π (D)
65π
68. If the p, q, r have truth values. F, F, T then the statement (p ↔ q) ∨ ~ r → (p ∧ r) will be (A) T (B) F (C) T, F (D) None
69. The statement (p ∨ q) ↔ (q ∧ ~ p) is a (A) Tautology (B) Contradiction (C) Neither tautology nor contradiction (D) None of these
70. If nth term of sequence 212 ,
1371 ,
911 ,
2320 , .... is
175
then value of n is - (A) 20 (B) 10 (C) 5 (D) 13 71. The ratio in which plane 2x – k = 0 divides the line
joining (–2, 4, 7) & (3, – 5, 8) is 9 : 1 then k equal to- (A) 4 (B) 5 (C) 6 (D) 7 72. If |a| = 2, |b| = 5 and |a × b| = 8 then |a – b| is equal to: (A) 12 (B) 15 (C) 17 (D) 5 73. The extremities of a line segment of length 6 move in
two fixed perpendicular lines. If locus of a point P which divides this line segment in ratio 1 : 2 is an ellipse then eccentricity of this ellipse is -
(A) 21 (B)
21 (C)
23 (D)
43
74. The value of p such that the vertex of parabola
y = x2 + 2px + 13 is 4 units above x-axis & lies in first quadrant is :
(A) 3 (B) 4 (C) ± 3 (D) – 3
XtraEdge for IIT-JEE APRIL 2010 75
75. If the lines represented by x2 + 2λx + 2y2 = 0 & lines represented by (1 + λ)x2 – 8xy + y2 = 0 are equally inclined then λ equals :
(A) – 2 (B) + 2 (C) ± 2 (D) ± 4 76. locus of centre of a variable circle tx2 + ty2 + 2(t2 + 1)x – 2(t2 – 1)y + t = 0 is a : (A) Straight line (B) Parabola (C) Ellipse (D) Hyperbola
77. If ∫−
4
1)x(f dx = 4 and ))x(f3(
4
2∫ − dx = 7 then the
value of ∫4
2)x(f dx is -
(A) 2 (B) – 3 (C) – 5 (D) None
78. Bisector of angle between lines 2x + y – 6 = 0 & 4x – 2y + 7 = 0 which contains origin is -
(A) acute angle bisector ; x = 5/8 (B) acute angle bisector ; y = 19/4 (C) obtuse angle bisector ; x = 5/8 (D) obtuse angle bisector ; y = 19/4
79. The value of [ ]∫2
1
1–))x(g(f f 'g(x). g'(x) dx where
g(1) = g(2) is equal to - (A) 1 (B) 2 (C) 0 (D) None
80. If f(x) = ∫ ++
2
22
x1xsinx sec2x dx and f(0) = 0 then
f(1) = (A) 1 – π/4 (B) π/4 – 1 (C) tan1 – π/4 (D) None of these
81. The domain of the function f(x) = ]x[x
xsec 1–
− is -
(A) R (B) R – (– 1, 1)I (C) R – I (D) R – [0, 1) 82. Let f(x) = sin–1x + sec–1x, then - (A) Df = [– 1, 1] (B) Rf = –π/2, π/2 (C) Rf = π/2 (D) None of these 83. If z1, z2, z3 represents the vertices of an equilateral
triangle such that |z1| = |z2| = |z3| then - (A) z1 + z2 = z3 (B) z1 + z2 + z3 = 0 (C) z1z2 = 1/z3 (D) z1 – z2 = z3 – z2
84. In a class of 100 students there are 70 boys whose average marks in a subject are 75.If the average marks of the complete class is 72, then what is the average of the girls.
(A) 73 (B) 65 (C) 68 (D) 74 85. A letter is taken at random from the letters of word
'STATISTICS' and a another letter is taken at random from letters of word 'ASSISTANT'. The probability that they are the same letter is -
(A) 1/45 (B) 13/90 (C) 19/0 (D) 5/18 86. If A = x : x ∈ I ; – 2 ≤ x ≤ 2 Β = x : x ∈ I ; 0 ≤ x ≤ 3 C = x : x ∈ N ; 1 ≤ x ≤ 2 and D = (x, y) ∈ N × N; x + y = 8 then - (A) n(A ∪ (B ∪ C)) = 5 (B) n(D) = 6 (C) n(B ∪ C) = 5 (D) None of these
87. Solution of sec2 ydxdy + 2x tan y = x3 is -
(A) tan y = 2xce− + (x2 – 1)
(B) tan y = 2xce− + (x2 – 1)
(C) tan y = 2xce− – (x2 – 1)
(D) None of these 88. The equation of common tangent to the curves y2 = 8x and xy = –1 is - (A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2 89. Let f be twice differentiable function such that f "(x) = – f(x) and f '(x) = g(x) h(x) = (f(x))2 + (g(x))2 . If h(5) = 11 then h(10) is equal to (A) 22 (B) 11 (C) 0 (D) None 90. We are required to from different words with the help
of letter of the word INTEGER. Let m1, be the number of words in which I and N are never together and m2 be the number of words which begin with
I and end with R. Then 2
1
mm is given by -
(A) 30 (B) 1/30 (C) 6 (D) 42
XtraEdge for IIT-JEE APRIL 2010 76
PHYSICS 1. If the amplitude of a damped oscillator becomes half
in 2 minutes, the amplitude of oscillation w.r.t. initial one after 6 minutes is
(A) 271 (B)
81 (C)
181 (D)
641
2. An infinite number of spring having force constants as k, 2k, 4k, 8k ..... ∞ and respectively are connected in series; then equivalent spring constant is
(A) k (B) 2k (C) k/2 (D) ∞
3. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m when the particle passes through the mean position. Its kinetic energy is 8 × 10–3 J. The equation of motion of this particle when the initial phase of oscillation is 45º can be given by
(A) 0.1 cos
π
+4
t4 (B) 0.1 sin
π
+4
t4
(C) 0.4 sin
π
+4
t (D) 0.2 sin
+
π t22
4. A mass m is moving with constant velocity along a line parallel to x-axis away from the origin. It angular momentum with respect to origin.
(A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing
5. A vessel containing oil (density = 0.8g/cm3) over mercury (density = 13.6 g/cm3) has a homogeneous sphere floating with half of its volume immersed in mercury and other half in oil. The density of material of sphere in g/cm3 is
(A) 3.3 (B) 6.4 (C) 7.2 (D) 2.8 6. Two trains move towards each other with the same
speed, speed of sound is 340 ms–1. If the pitch of the tone of the whistle of one is heard on the other changes by 9/8 times then the speed of each train is
V V (A) 2 ms–1 (B) 2000 ms–1 (C) 20 ms–1 (D) 200 ms–1
7. A sound level I differ by 4 dB from another sound of intensity 10 nW cm–2. The absolute value of intensity of sound level I in Wm–2 is
(A) 2.5 × 10–4 (B) 5.2 × 10–4 (C) 2.5 × 10–2 (D) 5.2 × 10–2
8. An ideal gas is taken through the cycle A → B → C → A as shown. If the net heat supplied to the gas in the cycle 5J, the work done by the gas in the process C → A
B
A
C2
1
10 P(N/m2)
V(m3)
(A) – 5 J (B) – 15 J (C) – 10 J (D) –20 J
9. There are n electrons of charge e on a drop of oil of density ρ. It is in equilibrium in an electric field E. Then radius of drop is
(A) 2/1
g4neE2
πρ
(B) 2/1
gneE
ρ
(C) 3/1
g4neE3
πρ
(D) 3/1
gneE2
πρ
10. Two identical cells of emf 1.5 V and internal resistance 1 Ω are in series. A third cell of similar parameters is connected in parallel to the
MOCK TEST – BIT-SAT
Time : 3 Hours Total Marks : 450 Instructions :
• This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical Reasoning (10) & English (15). There is Negative Marking
• Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.
• +3 Marks for each correct & – 1 Mark for the incorrect answer.
XtraEdge for IIT-JEE APRIL 2010 77
combination. The terminal voltage of the cells A, B, C are
A
1.5 V 1.5 V
1Ω B 1Ω
1.5 V 1Ω
C (A) 1, 1, 2 (B) 1.5, 1.5, 1.5 (C) 1.5, 0, 0 (D) 2, 1, 1
11. A wire has resistance of R ohm at T kelvin. At what temperature the resistance of wire be 2R ohm when temperature coefficient of resistance is α per degree centigrade.
(A) α
+α−2
1)T273( (B) α
−α−2
1)T273(
(C) α
−α− 1)T273( (D) α
+α− 12)T273(
12. Two cells each of same emf but of internal resistance r1 and r2 are joined to form a series circuit through an external resistance R. Value of R in term of r1 and r2 for which cell 1 has zero p.d. across it is
1
R
E E
r1 r2 2
(A) R = r1 – r2 (B) R = r1 + r2
(C) 21
21
rrrr
+ = R (D)
21
21
rrrr + = R
13. A current i flows in the network shown. Resulting mangnetic induction at point p is
2a
aA
B
a
a a
a
aF P
E D
C 2a
(A) a4iµ0
π (B) –
a8iµ2 0
π
(C) –aiµ
28 0
π (D)
aiµ
82 0
π
14. An alpha particle and a proton have same velocity when they enter a uniform magnetic field. The period of rotation of proton will be
(A) double of that of α particle (B) four times that of α particle (C) one half time that of α particle (D) same as that of α particle 15. A coil of inductance 8.4 mH and resistance 6 Ω is
connected to a 12 V battery. The current in the coil is 1A at approximate time
(A) 500 s (B) 20 s (C) 35 ms (D) 1 ms 16. A fish rising vertically up towards the surface with
speed 3ms–1 observe a bird diving vertically down towards it with speed 9 m/s. The actual velocity of bird is
y
y´
(A) 4.5 ms–1 (B) 5.4 ms–1
(C) 3.0 ms–1 (D) 3.4 ms–1
17. A concave lens of glass, refractive index 1.5 has both surface of same radius of curvature R. on immersion in a medium of refractive index 1.75, it will behave as a
(A) convergent lens of focal length 3.5 R (B) convergent lens of focal length 3R (C) divergent lens of focal length 3.5 R (D) divergent lens of focal length 3 R 18. I is the intensity due to source of light at any point P
on the screen if light reaches the point P via two different paths (a) direct (b) after reflection from a plane mirror then path difference between two paths is 3λ/2, the intensity at P is
(A) I (B) zero (C) 2I (D) 4I 19. The surface of some material is radiated, in turn, by
waves of λ = 3.4 × 10–7 m and λ = 5.4 × 10–7 m respectively. The ratio of stopping potential in two cases is 2 : 1, the work function is
(A) 2.05 eV (B) 1.05 eV (C) 3.05 eV (D) None 20. A X-ray tube has a working voltage of 40 × 103 V.
The continuous spectrum limit of the emitted x-rays is (A) 0.17 Å (B) 0.13 Å (C) 0.13 Å (D) 0.31 Å 21. The number of alpha and beta deca 88Ra222
experiences before turning into stable Pb206 isotope is (A) 4, 2 (B) 2, 4 (C) 1, 3 (D) 6, 10
XtraEdge for IIT-JEE APRIL 2010 78
22. The displacement of interfering light waves are y1 = 4 sin ωt and y2 = 3 sin (ωt + π/2). The amplitude of resultant wave is
(A) 5 (B) 7 (C) 1 (D) 0 23. A beam of light of wavelength 600 nm from a
distance source falls on a single slit 1 mm wide and resulting diffraction pattern is observed on a screen 2m away. Distance between first dark fringe on either side of the central bright fringe.
(A) 1.2 mm (B) 3.2 mm (C) 2.4 mm (D) 4.2 mm 24. The intensity of light from one source is double of
the other coherent source in a double slit experiment. The ratio of destructive to constructive interference in the obtained pattern is
(A) 34 (B) 1/34 (C) 17 (D) 1/17 25. Two radioactive material of half life T are produced
at different instants. Their activities area found to be A1 and A2 respectively when A2 < A1. Their age difference is
(A) 0.44 T log1
2
AA (B) 1.44 T log
2
1
AA
(C) 4.44 T log1
2
AA (D) 5.44 T log
1
2
AA
26. Three concentric conducting spherical shell x, y and z
have radii a, b and c respectively such that c > b > a, their surface charge density are σ, –σ and σ respectively. Then potential Vx is given by
C
b
a
y z
x
(A)
+−
εσ cb
ca 2
0 (B)
0εσ [a – b + c]
(C) 0ε
σ [a + b + c] (D) –0ε
σ [a + b – c]
27. A certain physical quantity is calculated from the
formula 3π (a2 – b2)h, where h, a and b are all lengths.
The quantity being calculated is (A) velocity (B) length (C) area (D) volume
28. The potential energy of a particle varies with distance
x from a fixed origin as U = Bx
xA2 +
, where A and B
are dimensional constant then dimensional formula for AB is
(A) M L7/2 T–2 (B) M L11/2 T–2
(C) M2 L9/2 T–2 (D) M L13/2 T–3 29 A particle leaves the origin at t = 0 and moves in the
+ve x axis direction. Velocity of the particle at any
instant is given by v = u
−
´tt1 . If u = 10 m/s and
t´ = 5 sec. Find the x coordinate of the particle at an instant of 10s.
t´ = 5 sec
t
v
u
(A) 0 (B) 10 m (C) 20 m (D) –10 m
30. An aero-plane drops a parachutist. After covering a distance of 40 m, he opens the parachute and retards at 2 ms–2. If he reaches the ground with a speed of 2ms–1, he remains in the air for about
(A) 16 s (B) 3 s (C) 13 s (D) 10 s
31. A tank moves uniformly along x-axis. It fires a shot from origin at an angle of 30º with horizontal while moving along positive x-axis & the second shot is also fired similarly that the tank moved along negative x-axis. If the respectively range of the shots are 250 m and 200 m along x-axis, the velocity of the tank.
30º
(A) 9.4 m/s (B) 4.9 m/s (C) 3.9 m/s (D) 5.9 m/s
XtraEdge for IIT-JEE APRIL 2010 79
32. A large number of particles are moving with same magnitude of velocity v but having random directions. The average relative velocity between any two particles average over all the pairs is
(A) 4π v (B)
2π v (C)
π3 v (D)
π4 v
33. A body is moving with uniform speed v on an horizontal circle from A as shown in the fig. Change in the velocity in the first quarter revolution is
v1
A
O
E
S
W
(A) v2 north (B) 2 v south west
(C) 2 v north-west (D) 2v west
34. A hollow vertical drum of radius r and height H has a small particle in contact with smooth inner surface of the upper rim at point P. The particle is given a horizontal speed u tangential to the rim. It leaves the lower rim at Q vertically below P. Taking n as an integer for number of revolution we get
Q
H
P u
(A) n = H
r2π (B) g/H2
r2π
(C) n = g/H2u
r2π (D) n = g/H2
r2uπ
35. A balloon is descending at a constant acceleration a. The mass of the balloon is M. When a mass m is released from the balloon is starts rising with acceleration a. Assuming that volume does not change when the mass is released, what is the value of m.
(A) )ga(
a2+
M (B)
+
a2ga M
(C) M)ga(
a2+
(D) ga
Ma+
36. Two blocks of masses 2 kg and 5 kg are at rest on ground. The masses are connected by a string passing over a frictionless pulley which in under the influence of a constant upward force F = 50 N. The accelerations of 5 kg and 2 kg masses are
5 kg 2 kg
50 N
(A) 0, 2.5 ms–2 (B) 0, 0 (C) 2.5 m/s2, 2.5 m/s2 (D) 1 m/s2, 2.5 m/s2
37. In the shown system m1 > m2. Thread QR is holding the system. If this thread is cut, then just after cutting
m1
m2
RQ
(A) Acceleration of mass m1 is zero and that of m2 is
directed upward (B) Acceleration of mass m2 is zero and that of m2 is
directed downward (C) Acceleration of both the block will be same (D) Acceleration of system is given by
+−
21
21
mmmm kg, when k is the spring factor
38. A car of mass M accelerates starting from rest.
Velocity of the car is given by v = 2/1
Mpt2
, where p
is the constant power supplied by the engine. The position of car as a function of time is given as
(A) 2/1
M9p8
t3/2 (B)
2/1
M8p9
t3/2
(C) 2/1
M9p8
t2/3 (D)
M8p9 t3
39. Six identical uniform rods PQ, QR, RS and ST, TV, UP each weighing w are freely joined at their ends to form a hexagon. The rod PQ is fixed in a horizontal position and middle points of PQ and ST are connected by a vertical string. The tension in string is
P Q
U R
T S (A) W (B) 3W (C) 2W (D) 4W
XtraEdge for IIT-JEE APRIL 2010 80
40. A body of mass 2 kg is being dragged with a uniform velocity of 2 ms–1 on a horizontal plane. The coefficient of friction between the body and the surface is 0.2. Work in 5 sec. is
(A) 39.2 J (B) 9.32 J (C) 23.9 J (D) 93.2 J
CHEMISTRY
1. 100 kg of iron ore (Fe2O3) containing 20% impurities on reduction with CO give iron equal to -
(A) 112 kg (B) 80 kg (C) 100 kg (D) 56 kg
2. Given : The mass of electron is 9.11 × 10–31 kg, Planck’s constant is 6.626 × 10–34 Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is -
(A) 5.79 × 107 ms–1 (B) 5.79 × 108 ms–1 (C) 5.79 × 105 ms–1 (D) 5.79 × 106 ms–1
3. The van der Waal equation for 0.5 mol of real gas is -
(A)
+ 2V4
aP
−
2bV = RT
(B)
+ 2V4
aP (2V – b) = RT
(C)
+ 2V4
aP = )b2–V(2
RT
(D)
+ 2V4
aP = )bV2(
RT2−
4. One mole of N2O4 is enclosed in a 5L container. At equilibrium, the container has 0.5 mol of N2O4. The equilibrium constant for the decomposition of N2O4
[N2O4 (g) 2NO2(g)] is-
(A) 1 (B) 52 (C) 3 (D)
51
5. Which one is the strongest Bronsted Lowry base out of the following -
(A) ClO– (B) −2ClO (C) −
3ClO (D) −4ClO
6. The pH of a solution obtained by mixing 50 mL of 0.4 M HCl with 50 ml of 0.2 N NaOH is -
(A) – log 2 (B) – log 2 × 10–1 (C) 1.0 (D) 2.0
7. Oxidation number of sulphur in H2SO5 is- (A) +2 (B) + 4 (C) +8 (D) +6
8. Equivalent mass of FeC2O4 in the reaction FeC2O4 → Fe3+ + CO2 is - (M.wt of FeC2O4 = M) (A) M (B) M/2 (C) M/3 (D) 2M/3
9. The fraction of total volume occupied by the atoms in a simple cube is -
(A) 4π (B) 2
8π
(C) 2 6π (D)
6π
10. In the diagram given below the value X is -
CuCuCu V50.0V15.02 → → ++++
E° = X Volt (A) 0.325 V (B) 0.65 V (C) – 0.35 V (D) – 0.65 V 11. For a first order reaction, A → B, t1/2 = 1 hr. What
fraction of the initial conc. of A reacts in 4 hrs ?
(A) 1615 (B)
161 (C)
87 (D)
81
12. An azeotropic solution of two liquids has boiling point lower than that of either of them if it -
(A) shows a –ve deviation from Raoult’s Law (B) shows a +ve deviation from Raoult’s Law (C) shows no deviation from Raoult’s Law (D) is saturated
13. In multi-molecular colloidal solution atoms or molecules are held together by -
(A) Hydrogen bonding (B) Strong attraction forces (C) Van der Waal’s forces (D) Strong electrical forces
14. Given : C + 2S → CS2 ; ∆H° = + 117 kJ C + O2 → CO2 ; ∆H° = –393 kJ S + O2 → SO2 ; ∆H° = – 297 kJ The value of ∆Hcombustion of CS2 in kJ mol–1 is (A) – 1104 (B) + 1104 (C) + 807 (D) – 807
15. Aspirin is chemically - (A) Methyl salicylate (B) Ethyl salicylate (C) Acetyl salicylic acid (D) o-hydroxy benzoic acid
16. Aniline when diazotized in cold and then treated with dimethyl aniline gives a colored product. It’s structure would be -
(A) (CH3)2N N = N
(B) (CH3)2N NH
(C) CH3NH N = N NHCH3
(D) CH3 N = N NH2
XtraEdge for IIT-JEE APRIL 2010 81
17. Ethyl ester Excess
MgBrCH3 → P. The product P will be -
(A) H3C
OH H3C
CH3
(B)
H3C
OH H5C2
C2H5
(C)
OH
C2H5 C2H5
C2H5
(D)
OH
C2H5H5C2
H7C3 18. When m-chlorobenzaldehyde is treated with 50%
KOH solution, the product(s) obtained is -
(A)
OH OH
CH – CH OH OH
(B)
OH OH
COO– +
CH2OH
(C)
Cl Cl
COO–
+ CH2OH
(D)
Cl Cl
CH – CH OH OH
19. Phenol is less acidic than - (A) Acetic acid (B) p-Nitrophenol (C) Both (A) & (B) (D) None of these
20.
OH
+ C2H5I OHHC.Anhy
HCO
52
52 → Product
In the above reaction product is - (A) C6H5OC2H5 (B) C2H5OC2H5 (C) C6H5OC6H5 (D) C6H5I 21. When acetylene is passed through dilute H2SO4
containing Hg2+ ions, the product formed is - (A) Acetone (B) Acetic acid (C) Acetaldehyde (D) Formaldehyde 22. Among the following statements on the nitration of
aromatic compounds, the false one is - (A) The rate of nitration of benzene is almost the same
as that of hexadeuterobenzene (B) The rate of nitration of toluene is greater than that
of benzene (C) The rate of nitration of benzene is greater than that
of hexadeuterobenzene (D) Nitration is an electrophilic substitution reaction
23. Which one is electrophilic addition - (A) CH3 – CH3 + Cl2 → C2H5Cl + HCl (B) CH3CH = O + HCN → (CH3)2C(OH)CN (C) (CH3)2C = O + HCN → (CH3)2C(OH)CN (D) CH3 = CH2 + Br2 → CH2BrCH2Br 24. Which one of the following conformation of
cyclohexane is chiral - (A) Twist boat (B) Rigid (C) Chair (D) Boat 25. The dipole moment is the highest for - (A) Trans-2-butene (B) 1,3- dimethyl benzene (C) Acetophenone (D) Ethanol 26. IUPAC name of the following compound -
N
CH3
CH3 C O
(A) N, N-dimethylcyclo propanecarboxamide (B) N-methylcyclopropanamide (C) Cyclopropanamide (D) None of the above 27. When a mixture of solid NaCl, solid K2Cr2O7 is
heated with conc. H2SO4 orange red vapours are obtained of the compound –
(A) Chromous Chloride (B) Chromyl Chloride (C) Chromic Chloride (D) Chromic sulphate
28. Which of the following will give a pair of enantiomorphs -
(A) [Co(NH3)4Cl2] NO2 (B) [Cr(NH3)6] [Co(CN)6] (C) [Co(en)2Cl2]Cl (D) [Pt(NH3)4] [Pt Cl6] 29. In following reaction
−4yMnO + xH+ + −2
42OC → yMn++ + 2CO2 + 2x H2O
x and y are (A) 2 and 16 (B) 16 and 2 (C) 8 and 16 (D) 5 and 2 30. A reduction in atomic size with increase in atomic
number is a characteristic of element of - (A) High atomic mass (B) d-block (C) f – block (D) Radioactive series 31. Which statement is not correct for nitrogen – (A) It has a small size (B) It does not readily react with O2 (C) It is a typical non-metal (D) d-orbitals available for bonding
XtraEdge for IIT-JEE APRIL 2010 82
32. Which one of the following is not an amphoteric substance -
(A) HNO3 (B) −3HCO (C) H2O (D) NH3
33. Which reaction cannot be used for the production of halogen acid –
(A) 2KBr + H2SO4 → K2SO4 +2HBr (B) NaHSO4 + NaCl → Na2SO4 + HCl (C) NaCl + H2SO4 → NaHSO4 + HCl (D) CaF2 + H2SO4 → CaSO4 + 2HF 34. B(OH)3 + NaOH NaBO2 + Na[B (OH)4] + H2O How can this reaction is made to proceed in forward
direction - (A) Addition of cis 1, 2-diol (B) Addition of borax (C) Addition of trans 1, 2-diol (D) Addition of Na2HPO4
35. Sodium thiosulphate is prepared by - (A) Reducing Na2SO4 solution with H2S (B) Boiling Na2SO3 solution with S in alkaline
medium (C) Neutralising H2S2O3 solution with NaOH (D) Boiling Na2SO3 solution with S in acidic medium
36. The critical temperature of water is higher than that of O2 because H2O molecule has -
(A) Fewer electrons than oxygen (B) Two covalent bond (C) V-shape (D) Dipole moment 37. Zone refining is a technique used primarily for which
one of the following process - (A) Alloying (B) Tempring (C) Sintering (D) Purification 38. Which one of the following elements has the highest
ionization energy – (A) [Ne] 3s2 3p1 (B) [Ne] 3s2 3p2 (C) [Ne] 3s2 3p3 (D) [Ar] 3d10 4s2 4p2
39. The correct order of dipole moment is - (A) CH4 < NF3 < NH3 < H2O (B) NF3 < CH4 < NH3 < H2O (C) NH3 < NF3 < CH4 < H2O (D) H2O < NH3 < NF3 < CH4 40. If Nx is the number of bonding orbitals of an atom
and Ny is the no. of the antibonding orbitals, then the molecule/atom will be stable if -
(A) Nx > Ny (B) Nx = Ny (C) Nx < Ny (D) Nx ≤ Ny
MATHEMATICS
1. Consider the sequence (angles are measured in radians) sin log102 , sin log103 , sin log104 ….then -
(A) all the terms of this sequence are positive (B) all the terms of this sequence are negative (C) 1001th term is negative (D) 10001th term is negative
2. The order relation between x, sin–1 x & tan–1x x ∈(0 ,1) is -
(A) tan–1x < x < sin–1x (B) sin–1x < tan–1x < x (C) x < sin–1x < tan–1x (D) None
3. The smallest positive valve of x satisfying the equation log2 cos x + log2 (1 – tan x ) + log2(1 + tan x) – log2sin x = 1 is -
(A) π/8 (B) π/6 (C) π/4 (D) π/6 4. A pole stands at a point A on the boundary of a
circular park of radius r and subtends an angle α at another point B on boundary. If arc AB subtends an angle α at the centre of the path, the height of the pole is -
(A) r sin α/2 tan α (B) 2r sin α/2 tan α (C) 2r sin α/2 cot α (D) None of these 5. The base of a triangle lies along the line x = a and is
of length 2a. The area of the triangle is a2. If the third vertex lies on the line -
(A) x = 0 (B) x = – a (C) x = 2a, or x = 0 (D) x = 0 or x = – 2a 6. If y = mx bisects an angle between the lines
ax2 – 2hxy + by2 = 0 then m
1–m2 =
(A) h
a–b (B) h
b–b (C) h
ba + (D) None
7. If the circle x2 + y2 + 2gx + 2fy + c = 0 passes through all the four quadrant then -
(A) g = – b (B) C > 0 (C) C < 0 (D) None
8. The equation of the circle which has two normals (x–1) (y – 2) = 0 and a tangent 3x + 4y = 6 is
(A) x2 + y2 – 2x – 4y + 4 = 0 (B) x2 + y2 + 2x – 4y + 5 = 0 (C) x2 + y2 = 5 (D) (x –3)2 + (y – 4)2 = 5
9. Circles drawn on the diameter as focal distance of any point lying on the parabola x2 – 4x + 6y + 10 = 0 will touch a fixed line whose equation is
(A) y = 2 (B) y = –1 (C) x + y = 2 (D) x – y = 2
XtraEdge for IIT-JEE APRIL 2010 83
10. The foci of a hyperbola coincide with the foci of the
ellipse 25x2
+ 9y2
= 1. If eccentricity of the
hyperbola is 2, then its equation is (A) x2 – 3y2 – 12 = 0 (B) 3x2 – y2 – 12 = 0 (C) x2 – y2 – 4 = 0 (D) None of these
11. →α and
→β are two mutually perpendicular unit vector
a→α + a
→β + c(
→α ×
→β ),
→α + (
→α ×
→β ) and c
→α + c
→β
+ b (→α ×
→β ) are coplaner then c is
(A) A.M. of a & b (B) G.M. of a & b (C) H.M. of a & b (D) None of these 12. The point of contact of the spheres x2 + y2 + z2 + 2x – 4y – 4z – 7 = 0 x2 + y2 + z2 + 2x – 4y – 16z + 65 = 0 (A) (1, 2, 6) (B) (1, 2, –6) (C) (1, –2, 6) (D) (–1, 2, 6) 13. If f(x) = 3 – 4x2 – 4x + 8–1 then range of f(x) is (A) (–∞, 1) ∪ (3, ∞) (B) (2, 3) (C) [2, 3] (D) None of these 14. If x > 0 and g is a bounded function then
1e
)x(ge)x(flim nx
nx
n ++
∞→ is
(A) 0 (B) f(x) (C) g(x) (D) None 15. If a1 = 1 and an = n(1 + an–1) ∀ n ≥ 2 than
+
+
+
∞→ n21n a11...
a11
a11lim =
(A) 1 (B) e (C) 1/e (D) None 16. Let f(x) = |2 sgn 2x| + 2 then f(x) has (A) removable discontinuity (B) infinte discontinuity (C) No discontinuity (D) essential discontinuity
17. If f(x) = cos
−
π 3x]x[2
, 1 < x < 2 and [.] = G.I.F.
then f´
π32
is
(A) 0 (B) 3(π/2)2/3 (C) –3(π/2)2/3 (D) None of these 18. If yex = cos x then, y4/ y = (A) –1 (B) 2 (C) –4 (D) None
19. Let f & g be differentiable function satisfying g´(a) = 2, g(a) = b and fog = I (Identity function), then f´(b) is equal to
(A) 1/2 (B) 2 (C) 2/3 (D) None
20. Tangents are drawn from origin to the curve y = sin x points of contact lie on the curve
(A) x2 + y2 = x2y2 (B) x2 – y2 = xy (C) x2 – y2 = x2y2 (D) None of these 21. Two positive numbers whose sum is 16 and sum of
whose cubes is maximum are given by (A) 8, 8 (B) no such number exist (C) 0, 16 (D) None of these
22. Let f(x) = 2x1 , g(x) =
x1 on [a, b], 0 < a < b. Let
)a(g)b(g)a(f)b(f
−− =
)c´(g)c´(f for same a < c < b then c is
(A) A.M. of a & b (B) G. M. of a & b (C) H.M. of a & b (D) None of these
23. ∫ + xcos1 2 . sin 2x cos 2x dx =
(A) 52 (1 + cos2x)3/2(3 – 2cos2x)2 + c
(B) 52 (1 + cos2x)3/2(3 – 2 cos2x) + c
(C) 52 (1 + cos2x)3/2(3 + 2 cos2x) + c
(D) None of these
24. ∫
+− ....
4x
2x1
42dx
(A) sin x (B) – sin x (C) cos x (D) None
25.
∫
∫
π
π
−
π→ π−
−
2
2
x
4/
x
2/
tcos
2x dt)2/t(
dt)12(lim =
(A) π
2loge (B) π22nl (C)
π2n2l (D) None
26. ∞→n
lim2n/2
2n11
+
2n/4
2
2
n21
+
2n/6
2
2
n31
+
....
2n/n2
2
2
nn1
+ =
(A) 4/e (B) 3/e (C) 2/e (D) None
XtraEdge for IIT-JEE APRIL 2010 84
27. The area bounded by the curves y = 2x2 & y = x
|x|
and x = 0 is equal to
(A) 32 (B)
622 (C)
62 (D) None
28. Order and degree of the differential equation "y = (y´+ 3)1/3 are respectively (A) 2, 2 (B) 2, 3 (C) 3, 2 (D) None
29. If x18 = y21 = z28 then 3, 3 logyx, 3 logz y, 7 logxz are in
(A) A.P. (B) G.P. (C) H.P. (D) None
30. If log2x + log2y ≥ 6 then least possible value of x + y is
(A) 32 (B) 16 (C) 8 (D) None
31. No. of real roots of the equation x3 + x2 + 10x + sin x = 0 is (A) 1 (B) 2 (C) 3 (D) ∞
32. The roots of the equation ax2 + bx + c = 0, a ∈R+ are two consecutive odd positive integers then
(A) |b| ≤ 4a (B) |b| ≥ 4a (C) |b| ≥ 2a (D) None of these
33. The sum of the terms of an infinitely decreasing G.P. is equal to the greatest value of the function f(x) = x3 + 3x – 9 on the interval [–2, 3] and the difference between the first two terms is f´(0) then sum of first terms is
(A) 19 or – 37 (B) 19 (C) –37 (D) None of these
34. If the complex number z1 = a + i, z2 = 1 + ib, z3 = 0 form an equilateral triangle (a, b are real number between 0 & 1) then :
(A) a = 3 – 1, b = 2/3
(B) a = 2 – 3 , b = 32 −
(C) a = 21 , b =
43
(D) None of these
35. ∑=
−n
0r
r)1( nCr
∞+++ .....
27
23
21
r3
r
r2
r
r is equal to
(A) 12
1n −
(B) 12
3n −
(C) 12
2n −
(D) None
36. The coefficient of x3y4z in the expansion of (1 + x + y – z)9 is (A) 2 . 9C7 . 7C4 (B) – 2 . 9C2 . 7C3
(C) 9C7 . 7C4 (D) None of these
37. If x1
ex
− = B0 + B1x + B2x2 + .... then Bn – Bn–1 = ?
(A) n1 (B)
n1 (C)
1n1−
(D) None
38. The number of point (x, y, z) in space whose each coordinate is a negative integer such that x + y + z + 12 = 0 is
(A) 55 (B) 110 (C) 75 (D) None 39. Six boys and six girls sit along a line alternatively
with probability P1 & along a circle (again alternatively) with probability P2 then P1/P2 is equal to
(A) 1 (B) 1/5 (C) 6 (D) None 40. If f(x) is a polynomial satisfying
f(x) = 21
−
x1f1
)x(fx1f)x(f
and f(2) = 17
then the value of f(5) is (A) 624 (B) –124 (C) 626 (D) 126
41. If A =
20x1
is idempotent then x =
(A) 0 (B) 2 (C) no such x exist (D) None of these 42. Let R be a relation on the set of integers given by
a R b if a = 2k b for some integer k then R is (A) an equivalence relation (B) reflexive and symmetric but not transitive (C) reflexive and transitive but not symmetric (D) symmetric and transitive but not reflxive
43. Minimum value of a
cb + + b
ac + + c
ba + , (for real
+ve numbers a, b, c) is (A) 1 (B) 2 (C) 4 (D) 6 44. From mean value theorem f(b) – f(a) = (b – a) f´(x1);
a < x1 < b if f(x) = x1 then x1 =
(A) ab (B) 2
ba + (C) ba
ab2+
(D) baab
+−
45. If f(x) = ∫ dxxcot4 + 31 cot3x – cot x and f
π
2=
2π
then f(x) is (A) π – x (B) x – π (C) x (D) None
XtraEdge for IIT-JEE APRIL 2010 85
LOGICAL REASONING
1. Fill in the blank spaces. 11, 12, 17, 18, 23, 24, (?) (A) 12 (B) 29 (C) 30 (D) 35
2. Choose the best alternative. Dum-Dum : Calcutta : : Palam : ? (A) Kerala (B) Delhi (C) Madras (D) Bombay
3. Pick the odd one out – (A) Wheat (B) Paddy (C) Towar (D) Mustard
4. Which of the following figures (A), (B), (C) and (D) when folded along the lines, will produce the given figure (X) ?
(X)
(A)
(B)
(C)
(D)
5. In each of the following questions, choose the set of figures which follows the given rule.
Rule : The series becomes complex as it proceeds :
(A)
(B)
(C)
(D)
6. In following question below, you are given a figure (X) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.
(X)
(A)
(B)
(C)
(D)
7. In the following question, find out which of the answer figures (A), (B), (C) and (D) completes the figure-matrix ?
?
(A)
(B)
(C)
(D)
8. The question that follow contain a set of three figures X, Y and Z showing a sequence of a piece of paper. Fig. (Z) shows the manner in which the folded paper has been cut. These three figures are followed by four answer figures from which you have to choose a figure which would most closely resemble the unfolded form of fig. (Z).
X Y Z
(A) (B)
(C) (D)
9. In the following question, complete the missing portion of the given pattern by selecting from the given alternatives (A), (B), (C) and (D).
(X)
?
(A)
(B)
(C)
(D)
10. In the following question, find out which of the answer figures (A), (B), (C) and (D) complete the figure - matrix ?
XtraEdge for IIT-JEE APRIL 2010 86
?
(A)
(B)
(C)
(D)
ENGLISH
1. Choose the one which best expresses the meaningful concept :
The state's duty is to . . . . the safety of its Citizens. (A) assure (B) ensure (C) insure (D) accept 2. Choose the one which best expresses the meaningful
concept : The company went . . . . in the 1990's recession. (A) burst (B) bust (C) bursted (D) busted 3. Choose the one which best expresses the meaningful
concept : What can we . . . . from this evidence, Watson ? (A) deduce (B) deduct (C) reduce (D) conduce 4. Choose the one which best expresses the meaningful
concept in opposite meaning : Zenith : (A) Sky (B) Firmament (C) nadir (D) naive 5. Pick up the correct Synonym for the following word Voracious : (A) Hungry (B) Wild (C) Quick (D) Angry 6. One who travels from place to place : (A) Journey man (B) Tramp (C) Itinerant (D) Mendicant 7. Choose the one which best expresses the meaning of
the given idiom/proverb : To fly off the candle : (A) To dislocate (B) To lose one's temper (C) To take off (D) To be indifferent 8. Fill in the blanks with one of the options given
below: Gandhi Ji . . . . . . smoking in his youth.
(A) took to (B) took for (C) took in (D) took up
9. Select the one which best expresses the same sentence in Passive or Active Voice.
Get the box broken. (A) Get someone to break the box. (B) They have broken the box (C) Have the broken box (D) Break the box
10. Choose the one which best expresses the correct answer in the speech :
He said, "How shabby you are looking!" (A) He asked how shabby I was looking (B) He exclaimed with disgust that I was looking
very shabby (C) He exclaimed with sorrow that they were
looking much shabby (D) He told that I was looking much shabby
11. Pick out the mis-spelt word – (A) Neigh (B) Rein (C) Neice (D) Neither
12. Find out which part of the sentence has an error :
)a(wonderI /
)b(bookthewithdonehashewhat /
)c(himlendI /
)d(ErrorNo
(A) Wonder (B) What he has done with the book (C) I lend him (D) No Error
13. Pick out the most appropriate pair to fill in the blanks in the same order, to make the sentence meaningfully complete :
She was . . . . . because all her plans had gone . . . . . (A) distraught, awry (B) Frustrated, Magnificently (C) Elated, wild (D) Dejected, splendidly 14. Pick out the most effective word from the given
words to fill in the blanks to make the sentence meaningfully complete :
Most of the issues discussed in the meeting were trivial and only a few were :
(A) Interesting (B) Practical (C) Complex (D) significant 15. Pick out the most appropriate pair to fill in the
blanks in the same order, to make the sentence meaningfully complete :
The . . . . . of glory lead but to the . . . . . (A) Paths, grave (B) Ways, happiness (C) Acts, Prosperity (D) Achievements, Suffering
XtraEdge for IIT-JEE APRIL 2010 87
CHEMISTRY
1.[A]
C6H5 – C||O
C||O
CH3 H3C
O||C O
||C
–C6H5 Orientation for aldol condensation
2.[B] A →Ozonolysis B ∆
→ 422 SOH,OHNH oxime
formation followed by Beckmann rearrangement.
O||C –NH–CH3
N-methyl benzamide
So, B is
3CH|
OC = and
A is
33 CHCH||CC =
3.[D] (CH3)3CCH2CH2OH
Heat,OHH,OCrK
2
722 →+
(CH3)3)A(
2CCH COOH
→ 2SOCl (CH3)3 CCH2COCl(B)
(CH3)2NH
(CH3)3)C(
2CCH C||O
N(CH3)2
(CH3)3)D(
CC H2CH2N(CH3)2
OHether,LiAlH
2
4 ←
4.[C] Attack of nucleophile is a rate determining step
F3C
O
O
O
I II
CH3
I is more electron deficient and facilitates a faster attack.
5.[A] As water introduces, water dissolves HCl(g) and a press drop is produced Liquid level in the capillary rises.
6.[D] 7.[B] The energy of AOs depends on the (n + l) values n + l value of (n – 1) d = n + 1 ; n + l value of ns = n (n + l) value of (n + 1)d = n + 3 ; n + l value of nf
= (n + 3) But due to lower value of principle quantum no.
energy of nf < (n + 1) d ∴ energy of (n + 2) s < nf. 8.[A] A, B and C are magnesium ,aluminium and
silicon. Magnesium form ionic oxide, MgO ; Aluminium forms amphoteric oxide ,Al2O3 and silicon forms a giant molecule SiO2.
9.[ B,C,D] Due to resonance cyclohexatriene cation is
aromatic which causes it's stability. 10.[ A,B,D] Catalyst lower the activation energy of
forward reaction & backward direction keeping the same enthalpy of reaction.
11.[ A,B,C] Greater the value of (IE – ∆eg H) greater is the
electronegativity E.N. of P = (1680 + 340)/(4.18) (125) ~– 4 12.[A,B,C] H = E + PV
TdP
dH
=
TdPdE
+ P
TdPdV
+ V
For liquid, TdP
dH
=
TdVdE
TdPdV
+
P TdP
dV
+ V
For incompressible liquid, TdP
dV
~– 0.
∴ TdP
dH
~– V
SOLUTION FOR MOCK TEST IIT-JEE (PAPER - I)
XtraEdge for IIT-JEE APRIL 2010 88
For ideal gas, TdP
dH
= 0
For the real gas, if TdV
dE
= 0,
then TdP
dH
= P
TdPdV
+ V ≠ 0
13.[A] F(Monoester) Molecular weight = 186 No Br2 reaction ⇒ Saturated Two oxygen ⇒ 2 × 16 = 32 No. of CH2 = (186 – 32)/14 = 11 Hence, molecular formula of saturated monoester
F, is C11H22O2 Hydrolysis
G H Optically active, soluble in NaOH ⇒ Acid We have, Ag+ salt →2Br ± J Hunsdiecker reaction; radical intermediate, so racemic mixture (J contains one carbon less than G)
Optically active, (Alcohols are not soluble in NaOH) +ve iodoform test, it suggests HO–
3CH|
R–CH On warning with H2SO4 (dehydration)
I (no diastereomers means no geometrical isomers) It suggests same alkyl group on one of the doubly bonded carbon atoms
H NaBr.IITsCl.I optically active ⇒ J
It suggests G contains one more carbon atom than H.
Hence, molecular formula of ester F, is
C5H11– C||O
–O–C5H11
C5H11– C||O
–OH(G) C5H11OH(H)
⇒ HO – CH CH3
* – –CH|
CH3
CH3 →∆,SOH 42
C=C
H
H3C (I)
No cis-trans isomers
CH3
CH3
H NaBr
TsCl → CH3–
)J(33
*
CHCH||
Br–HC–CH
So, G is CH3–
333 CHCHCH|||
OHCCHCH||O
−−−
)J(
3333
3Br*
3
CHCHCHCH||||
BrCHCHCHOAg
O||CHCCHCH 2
±
−−−→−−−−
F = CH3–
3333
3
CHCHCHCH||||
CHCHCHOCCHCH||O
−−−−−−
14.[C] I exists as diastereomers and H is optically active. So, H is HO–
3CH|CH − CH2CH2–CH3
H
H3CC = C
H
CH2CH3
H2SO4 ∆
, and
G is CH3CH2CH2–
3CH|
OHCHC||O
−
15.[D] Since H is optically active and gives negative iodoform test, so H is
HO–CH2–
3CH|
–CH CH2–CH2 and
G is CH3CH2
3
2
CH|CHCH C
||O
–OH
16.[A]
17.[D] Final pH = 1.7 or – log [H+] = 1.7 or log [H+] = – 2 + 0.3 or [H+] = 2 × 10–2 Let v ml of 0.1 (M) HA solution is mixed with
100 ml of 10–2 (M) HCl. In the mixed solution,
[HA] = )v100(
10v 4
+× −
× 103 =)v100(
v1.0+
(M) and
[HCl]=)v100(
0110 33
+×−
= )v100(
1+
HA H+ + A–
)v100(v1.0+
(1 – α) )v100(
v1.0+
α )v100(
v1.0+
α
+)v100(
1+
∴ [H+] = )v100(
)1v1.0(+
+α = 2 × 10–2
XtraEdge for IIT-JEE APRIL 2010 89
or 0.1 v α + 1 = 2 + 2 × 10–2 × v
Ka =
)v100()1(v1.0
)v00.1(v1.0
)v100()1v1.0(
+α−
+α
×+
+α
=)1( α−
α × 2 × 10–2
∴ )1( α−
α × 2 × 10–2 = 10–2 or α−
α1
= 21
or 2α = 1 – α or α = 1/3
∴ 0.1 v × 31 + 1 = 2 + 2 × 10–2 × v
or 30v –
50v = 1 or
5030v20
× = 1
or v = 2
150 = 75 ml
18.[D] For isohydric solution, Ka1C1 = Ka2C2 Column Matching 19. [A] → r,s,t; [B] → p,r,s; [C] → s; [D] → q,t
For CH3CH2CH2NH2 →HCl/NaNO2
productthenot;ermediateintsimply
isionDiazoniumNCHCHCH 2223+
CH3–CH2–CH2–OH
N – CH3 + HO CH3
CH3
CH3
on heating doesn't give alkene.
20. [A] → q,r; [B] → p,r; [C] → r,s; [D] → p,t
C6H5CH2–CHO →−
2Br/OH C6H5–
)(Br|
CH
±
–CH = O
CH3CHO →−
−OH
Cº250CH3–
)(CH|
OH
±–CH2–CH = O
CH3–CH2CH = O Cº25OH →
−CH3CH2–CH=
3CH|
CHOC −
CH3–
3CH|
CH − CH = O →−OH.Conc
CH3–
3CH|CH − C
||O
–O– + CH3–
3CH|CH − CH2OH
MATHEMATICS
1.[A] Let us first count the number of elements in F.
Total number of functions from A to B is 34 = 81. The number of functions which do not contain
x(y) [z] in its range is 24. ∴ the number of functions which contain exactly
two elements in the range is 3 . 24 = 48. The number of functions which contain exactly
one element in its range is 3. Thus, the number of onto functions from A to B is
81 – 48 + 3 = 36 [using principle of inclusion exclusion] n (F) = 36. Let f ∈ F. We now count the number of ways in
which f –1(x) consists of single element. We can choose preimage of x in 4 ways. The
remaining 3 elements can be mapped onto y, z is 23 – 2 = 6 ways.
∴ f –1 (x) will consists of exactly one element in 4 × 6 = 24 ways.
Thus, the probability of the required event is 24/36 = 2/3
2.[A] Let E1 denote the event that the letter came from
TATANAGAR and E2 the event that the letter came from CALCUTTA. Let A denote the event that the two consecutive alphabets visible on the envelope are TA. We have P(E1) = 1/2, P(E2) = 1/2, P(A / E1) = 2/8, P (A / E2) = 1/7. Therefore, by Bayes' theorem we have
P(E2 / A) = )E/A(P)E(P)E/A(P)E(P
)E/A(P)E(P
2211
22
+
= 114
3.[D] Required probability = 1 – P (all the letters are
put in correct envelops) The number of the ways of putting the letters in
the envelops = 4P4 = 4! The number of ways of putting letters in correct
envelops = 1
∴ Required probability = 1 – 241 =
2423
4.[D] We have
AB =
− x52x10001000x5
=
100010001
⇒ x = 1/5
5.[B] Greatest term in the expansion of (x + y)n is
kth term where k =
++
yxy)1n(
XtraEdge for IIT-JEE APRIL 2010 90
In the present case
k =
++
x23)x2)(150( =
+ 5/23)5/2)(51( =
17102 = 6
Thus, 6th term is the largest term.
6.[D] We have | z | = z4
z4z +− ≤
z4z − +
|z|4
= 2 + |z|
4
⇒ | z |2 ≤ 2 | z | + 4 ⇒ (| z | – 1)2 ≤ 5 ⇒ | z | – 1 ≤ 5 ⇒ | z | ≤ 5 + 1
Also, for z = 5 + 1
z4z − = 2
Therefore, the greatest value of | z | is 5 + 1.
7.[D] Integrating by parts, the given integral is equal to
x tan–1 16
11x − ∫
−−
16
1 1xx4
1x
x dx
= π3
16 – ∫−
16
1 1x
dx41
= π3
16 – ∫+
3
0
2dt
t)t1(t4
41 ( x = 1 + t2)
= π3
16 – ( )33 + = 3
16π – 2 3
8.[C] The intersection of y – x + 1 = 0 and y + x + 5 = 0 is (– 2, –3). Put x = X – 2, y = Y – 3. The given
equation reduces to dXdY =
XYXY
+− . This is a
homogeneous equation, so putting Y = υX, we get
XdXdυ =
112
+υ+υ
−
⇒
+υ−
+υυ
−1
11 22 dυ =
XdX
⇒ –21 log (υ2 + 1) – tan–1 υ = log | X | + C
⇒ log (Y2 + X2) + 2 tan–1 XY = C
⇒ log ((y + 3)2 + (x + 2)2) + 2 tan–1 2x3y
++ = C
9.[A] We have (1 + x)n – nx – 1 = C0 + C1 x + C2 x2 + … + Cn xn – nx – 1 = x2 [C2 + C3x + … + Cn xn–2] [Q C1 = n, C0 = 1] Thus, (1 + x)n – nx – 1 is divisible by x2.
10.[A,B,C,D] y = 1)A2sinA2(cos
1)A2sinA2(cos2
2
−−
+−
⇒ y = 1)A2sinA2(cos1)A2sinA2(cos
−−±+−±
which gives us four values of y, say y1, y2, y3 and y4. We have
y1 = 1A2sinA2cos1A2sinA2cos
−−+− =
A2sin)1A2(cosA2sin)A2cos1(
+−−+
= AcosAsin2Asin2
AcosAsin2Acos22
2
+−−
= )AsinA(cosAsin)AsinA(cosAcos
−− = cot A
y2 =1)A2sinA2(cos1)A2sinA2(cos
−+−+−− =
A2sin)A2cos1(A2sin)A2cos1(
−+−+−
=AcosAsin2Acos2
AcosAsin2Asin22
2
−−
+ = – tan A
y3 =1)A2sinA2(cos
1)A2sinA2(cos−+−
+− =A2sin)A2cos1(
A2sin)A2cos1(−+−
−+
=AcosAsin2Acos2
AcosAsin2Acos22
2
−−− = –
AsinAcosAsinAcos
+−
= –Atan1Atan1
+− = – tan
−
π A4
= – cot
+
π A4
y4 =1)A2sinA2(cos1)A2sinA2(cos
−++−− =
A2sin)A2cos1(A2sin)A2cos1(
+−−+−
=AcosAsin2Asin2
AcosAsin2Asin22
2
+−+ =
AsinAcosAsinAcos
−+
=Atan1Atan1
−+ = tan
+
π A4
.
11.[B, C, D] Equations of the given circles can be written as (x – 3)2 + y2 = 32 (1) and (x + 1)2 + y2 = 12 (2) Equation of any tangent to circle (2) is (x + 1) cos θ + y sin θ = 1 (3) This will be a tangent to circle (1) also if
θ+θ
−θ+22 sincos
1cos)13( = ± 3 ⇒ 4 cos θ – 1 = ± 3
That is, cos θ = 1 or cos θ = –21 . When cos θ = 1,
we have sin θ = 0, and the equation of the common tangent (3) becomes
x + 1 = 1 or x = 0 (4) When cos θ = –1/2, we have sin θ = ± 2/3 , and
the equations of the common tangents are
–21 (x + 1) ±
23 y = 1 ⇒ x – y3 + 3 = 0 (5)
and x + y3 + 3 = 0 (6)
XtraEdge for IIT-JEE APRIL 2010 91
12.[A, B, C, D]
Let F(x) = ∫ ++−2x
0 t
2
e24t5t dt
⇒ F′(x) = 2x
24
e2
4x5x
+
+− .2x
So from F′(x) = 0, we get x = 0 or
x2 =2
16255 −± =2
35 ± = 4, 1
Hence x = 0, ±2, ±1. 13.[B]
D C
–1 1 x = 4 x = –4
y = –4
y = +4
A B
Shaded region is S0. Area of S0
= 4 × 2 – 21 π (1)2 = 8 – π/2
14.[D] y ∈ [0, 4], x ∈ [–1, 1]
m (t) = cost
lines y = 2x + 0.4 lies inside the region so
⇒ t ∈ [0, 1]
t2 + (2t + 0.4)2 – 1 ≥ 0 ⇒ t ∈ [0.28, 1]
15.[B] (Slope)max. = (cos t)max = cos (0.28) and point is (π, 1)
π−
−x
1y = cos (0.28)
16.[C] SCSC
2
1 =2
1
rr =
13
x =13
)3(1)1(3−−− =
26− = – 3
17.[D] tangents = y = ±3
x + RT
tan30º =3
RT ⇒ RT = 3
18.[A]
•2
C1
C2 60º 1
(h + 1)2 + k2 = (1 + 2)2 (circle)
Column Matching 19. [A] → r; [B] → p,r; [C] → s; [D] → r
(A) ∑
=
10
0rr
20 C = 20C0 + 20C1 +……+ 20C10
But, 20C0 + 20C1 +……+ 20C20 = 220 Also, 20C20 = 1 = 20C0, 20C19 = 20C1, 20C18 = 20C2 etc. ∴ given sum = (20C0 + 20C1 +……+ 20C20) – (20C11 +…..+ 20C20) 220 + 20C10– (20C10 + 20C9 + ……+ 20C0) ∴ 2 (20C0 + 20C1 +…..+ 20C10) = 220 + 20C10
(B) ∑=
100
0rr
100 C (x –3)100–r 2r
= ((x–3) +2)100 = (x –1)100 = (1 –x)100
∑=
−100
0r
rr
100 )x(C = ∑=
−−100
0r
rr
100rr xC)1()1(
∴ Coeff. of x53 = (–1)53 100C53 = – 100C53
(C) We have (1+ x)10 = 10C0 + 10C1 x +10C2x2 +……+ 10C10 x10
....(1) Also (1–x)10 = 10C0 – 10C1x + 10C2x2 +……. …..+ 10C10x10 ....(2) Multiplying, we get (1 –x2)10 = (10C0 + 10C1 x + 10C2x2 +…… ….+ 10C10x10) × (10C0 –10C1x + 10C2x2+… …...+ 10C10 x10)
Equating the coefficients of x10, we get 10C5 (–1)5 =10C0
10C10 –10C1 10C9 + 10C2 10C8 +… ….+ 10C10 10C0 ⇒ – 10C5 = (10C0)2 – (10C1)2 + (10C2)2 +…… …+ (10C10)2
XtraEdge for IIT-JEE APRIL 2010 92
(D) 95C4 + ∑=
−5
0j3
j100 C
= 95C4 + 99C3 + 98C3 + 97C3 + 96C3 +95C3 = (95C4 + 95C3) +96C3 + 97C3 + 98C3+ 99C3 = (96C4 + 96C3) + 97C3 +98C3 +99C3 = (97C4 +97C3) +98C3 + 99C3 = (98C4 + 98C3) + 99C3 = 99C4 + 99C3 = 100C4
20. [A] → p,q; [B] → q,t; [C] → q; [D] → s (A) Given lines intersect if
12
11544312
λλ−−−
= 0
⇒ λ = 0, – 1
(B) ∞→x
lim 4x
++
+
++
−−
2x1x1
2x1x1
tan 1 =2 = y2 + 4y + 5
⇒ y = –1, – 3 (C) y2 – ax (– x – y) = 0 ⇒ for perpendicular lines a + b = 0 ⇒ 1 + a = 0 ⇒ a = – 1 (D) ( a × b ) × a = ( j – k ) × a ⇒ ( a . a ) b – ( a . b ) a = ( j – k ) × a on solving,
we get b = i
PHYSICS
1.[B] T = m1r1ω12
also T = m222
22
2 rg ω+
∴ m1r1ω12 = m2 2
222
2 )r(g ω+
or, 0.1 × ω12 =
21 22 )10()10( +
0.1 ω12 = 1
ω1 = 10 rad/s v1 = r1ω1 = 10 m/s 2.[A] As f = µN = mg
or, µmlω2 = mg ⇒ ω = lµ
g
3.[C] At terminal velocity net force is zero.
6πη(r1 + r2) VT +34
π (r13 + r2
3) ρg =34
π (r13 + r2
3) σg
4.[C] As supporting plane is lowered slowly ∴ N = mg – kx
5.[D] Net force acting on container due to liquid coming out from the holes is given by
F = ρA
×−×
4Hg2
4H3g2 = ρgAH towads left
∴ F = f = ρgAH towards right.
Now, τF = ρgAH × 2H into the plane of paper.
τf = ρgAH × 2H out of plane of papers
∴ τF = τf hence τN = 0
6.[C] TA = Rn
VP
A
AA and TB = Rn
VP
B
BB
Given, PA = PB , VA = VB and nA = 2nB
∴ TA = 2
TB
Now, A
B
B
A
B
A
MM
TT
VV
×= = 2
7.[A] Equivalent circuit
2, 2'
V0
ε2 C2V0
C1V0
ε1
1 1'2
2'3 3'
Middle plate
11' 33'
Total charge on 2 & 2' plate = Vd
Ad
A
2
02
1
01
εε+
εε
σ = ε0V
ε+
ε
2
2
1
1
dd
8.[A]
A
30B
15
30ºC
15º
O
AI
30º 120º
45ºC
BR O
l
Sine rule
XtraEdge for IIT-JEE APRIL 2010 93
º45sin
R = º30sin
l
l
R = 2º30sinº45sin
=
9.[A,B] As aT = aN
∴ ds
vdv = Rv2− which can also
be written as dtdv = –
Rv2
Integrating the above equations answer is obtained.
10.[A,C] (A) ρBLAg = ρ × 4L Ag + 2ρ ×
4L Ag
(B) FB = ρ × 4L Ag + 2ρ ×
4L3 Ag
FB = 47 ρLAg; a =
mmgFB −
11.[B,D]
c 6V
b 3Ω
8V
a
10
13V
•
11
← eq. ckt.
Using Kirchoff's Law Solve the circuit.
12.[A,B,C]
V
CV –CV
If battery is disconnected and plate are pulled
apart, then charge will remain constant
E = 0A2
Q∈
× 2 = 0
Q∈Α
∴ E remain same (A) is correct work is done against attractive force
→ ←−+ Fext.Felc by Fext. (B) is correct.
U = 21 CV2
V = constant [as battery is connected]
C = d
A0∈
as d increase C decrease ∴ U decrease option (C) is correct. 13.[C] As springs are in parallel
∴ a = )mm(x)kk(
massF
21
21net
++
=
14.[A] Frictional force on m2 will act in direction of
displacement if k2x > m2 a
15.[A] as k2A – µm2g = m2amax
∴ k2A – µm2g = m2
++
21
21
mmkk A
Solve to get answer. ]
16.[C] Induced emf across OP = 2
2B
21
ωl =
8B 2lω
(i) current = R8
B 2lω … (i)
Torque on the rod = 2 ∫2/
0
dxxBil
= 4
Bi 2l …(ii)
∴ R32
Bdtd
12M 422 ll ω
−=ω
× [substituting τ = I α]
– RM8B3d 22
0
l=
ωω
∫ω
ω∫t
0
dt
Solving this eq. & eq. (i)
i = R8
B 20lω e–αt
17.[B] θ = ∫∞
0
dti
18.[A] Heat generated = 20I
21
ω
Column Matching 19. [A] → p; [B] → p; [C] → r; [D] → q As cube is floating ρsALg = ρLAxg
∴ x =
ρρ
L
s L
20. [A] → r; [B] → p; [C] → s; [D] → q
S = 21 × 2 × 16 = 16 m
| Wg | = mg S = WN = m(g + a) cos2θ. S Wf = m(g + a) sin2θ . S
XtraEdge for IIT-JEE APRIL 2010 94
XtraEdge for IIT-JEE APRIL 2010 95
CHEMISTRY
1.[C]
NH2
NOBr
NH2 +
Br–
Br
+ Enantiomer
2.[B]
CH2=C
COOH
Ph H2/Pd
CH3–CH COOH
Ph (resolvable)
'A'
'B'
CH=CH–COOH H2/Pd
Optically inacitve
CH2–CH2–COOH
3.[B] Distribution of electrons in the MO's in He2 is σ1s
2 σ∗1s2 . He2 is unstable
Distribution of electrons in the MOS in H2 is σ1s2
. H2 is stable. 4.[B] After mixing total moles of A– = 100 × 0.2 × 10–3 + 100 × 0.3 × 10–3
= 100 × 10–3 × 0.5 moles After mixing total moles of HA = 100 × 0.1 × 10–3 + 100 × 0.2 × 10–3
= 100 × 0.3 × 10–3 moles
After mixing resulting pH = 5 + log35
5.[A,B,C]
CH2–CH*
NH2
COOH HO HO
It contains 2 phenolic hydrogens and a carboxylic acidic hydrogens
CH2–HC NH2
COO– HO HO
+
Zwitter ion
6.[A,B,C]
Grignard reagent reacting with acyl halide usually gives 3º alcohol.
7.[B,C,D] The order of H-bond energies F – H …….. F– > F – H …….. O > F – H …….. F >
O – H …….. O > O – H …….. F > N – H …….. N
8.[C,D]
9.[C,D] RHE reaction : Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl– LHE reaction : 2Ag(s) + 2Cl– → 2AgCl(s) + 2e Net reaction : Hg2Cl2 (s) + 2Ag(s) → 2Hg (l) + 2AgCl(s) In case of same concentration of Cl– ions in the
two half cells, Ecell is independent on the concentration of Cl–. Other substances are either pure solids or liquids, which have unit activities irrespective of their amounts.
Column Matching : 10. [A] → p, r, s ; [B] → q;
[C] → p, q,,s ; [D] → q
11. [A] → p, q,t ; [B] → p, r, s; [C] → p, q, r; [D] → p, q, r
Numerical Response type questions : 12. [4]
EW =
96500Q or
n/4.106977.2 =
96500606013 ×××
∴ n = 4
13. [8]
O
)reductionKishnerWolff(
glycol/KOH
NHNH 22
−
→
(W)
)Ozonolysis(
OH/O 23 →
O
OHO
HO(X)
SOLUTION FOR MOCK TEST
PAPER - II IIT-JEE (PAPER - II)
XtraEdge for IIT-JEE APRIL 2010 96
∆
→ 2)OH(Ca O
(Y)
)oxidation
VilligerBaeyer(
COOOHHC 56
− →
(Z)
O
O
14. [1] Molecular weight of starch = 162n ∴ moles of C6H12O6 to be produced = 1 mol
∴ moles of ATP required =30
3000 = 100 mol
15. [7]
0
0
PPP − = x1 = mole fraction of solute
x1 = 0.0125
−1
x1
1=
18m1000
×
∴ m = 0.70
16. [5] Eq. of metal = Eq. of hydrogen
X/511.0 =
112009.43 ⇒ X = 2
Now eq. of metal = eq. of KMnO4
2Y/51
1.0−
=1000
8.581.0 ×
∴ Y = 5
17. [7] For the first bulb p1v1 = n1RT & second bulb p2v2 = n2RT or p1v1 + p2v2 = (n1 + n2) RT …(1) Suppose equilibrium pressure at each bulb = p atm Then p(v1 + v2) = (n1 + n2)RT …(2) From eq. (1) & (2) p1v1 + p2v2 = p(v1 + v2) 9 × 5 + 6 × 10 = p × 15 p = 105/15 = 7 18. [2]
Xe
• •
O O
O
XeO3
Xe
• • F
O FF
F
XeOF4
19. [6] Each has two geometrical isomers and two optical isomers (shown by cis-isomer).
MATHEMATICS
1. [A] We have
∑ ∑ ∑= = =
−=−n
1r
n2
1r
n
1r
444 )r2(r)1r2( = f(2n) – 16f(n)
2. [D] Using De Moivre's theorem fr (α) =
2r/ie α 2r/i2e α …. r/ie α
= )r......21)(r/i( 2e +++α
= ]2/)1r(r)[r/i( 2e +α = )r/11)(2/i(e +α
∴ )(flim nnπ
∞→=
∞→nlim )n/11(2/ie +π
= eiπ/2 = cos
π
2 + i sin
π
2 = i
3. [D] z = tt
δ+γβ+α ⇒ ( γ + δt) z = α + βt
⇒ (δz – β)t = α – γz
⇒ t = β−δ
γ−αz
z [Q αδ – βγ ≠ 0 ]
As t is real, β−δ
γ−αz
z = β−δ
γ−αz
z
⇒ (α – γz)( zδ – β ) = ( α – zγ )(δz – β)
⇒ ( γ δ – γ δ )z z +(γ β – α δ)z + (α δ – β γ ) z
= (α β – α β) ...(1)
Since δγ is real,
δγ =
δγ or γ δ – δ γ = 0
Therefore (1) can be written as a z + a z = c ...(2) where a = i(α δ – β γ ) and c = i( α β – α β ) Note that a ≠ 0 for if a = 0 then
a δ – β γ = 0 ⇒ βα =
δγ =
δγ [Q
δγ is real]
⇒ αδ – βγ = 0, which is against hypothesis. Also, note that c = i( α β – α β ) is a purely real
number.
Thus, z = tt
δ+γβ+α represents a straight line.
4. [C] dxdy = (x – 1)(x – 2)2 so 2
2
dxyd = (x – 2) × (3x – 4).
The points of inflection are given by 2
2
dxyd = 0
So x = 2, x = 4/3 are points of inflection.
XtraEdge for IIT-JEE APRIL 2010 97
5.[A,B,D] We have adj A = |A| A–1 adj (AB) = |AB| (AB)–1 = |A| |B| (AB)–1 = |A| |B| (B–1 A–1) = ( |B| B–1) (|A| A–1) = (adj B) (adj A) 6.[A,B,C] We have
A(α, β)′ =
ααα−α
βe000cossin0sincos
=
α−α−−α−α−
βe000)cos()sin(0)sin()cos(
= A (–α, β) Also, A (α, β) A( – α, –β)
=
αα−αα
βe000cossin0sincos
ααα−α
β−e000cossin0sincos
= I
⇒ A(α, β)–1 = A(–α, –β) Next, adj A(α, β) = |A (α, β)| A (α, β)–1 = eβ A(– α, – β).
7.[A,B,C,D]
Let (r + 1) th term of n
42 x
x5
+ be
independent of x. We have
Tr + 1 = nCr
rn
2x5 −
(x4)r
= nCr 5n – r x6r – 2n For this term to be independent of x, 6r – 2n = 0
or n= 3r. As each of 18, 21, 27 and 99 is divisible by 3, each of this can be a possible value of n.
8.[A,B] Let k = 2n + 1, then 2n + 1Cr is maximum when
r = n. Also 2n+1Cn = 2n + 1Cn+1 Thus, kCr is maximum when
r = 21 (k – 1) or r =
21 (k + 1).
9.[A,B,C,D] As a912, a951 and a480 are divisible by 3, none of
them is prime. For a91, we have
a91 = 43421
times91
)9....99(91 =
91 (1091 – 1)
= 91 [(107)13 – 1]
=
−−110
1)10(7
137
−−1101107
= [(107)12 + (107)11 + ….+ 107 + 1] × [106 + 105 + ….. + 10 + 1] = a91 is not prime.
Column Matching : 10. [A] → p, r; [B] → p,q, r; [C] → t ; [D] → p, q, r
sin θ = 23 = sin
3π
⇒ θ = nπ + (–1)n 3π = 2nπ +
3π
4 sin θ cos θ – 2 sinθ – 2 3 cos θ + 3 = 0
⇒ (2 sin θ – 3 ) (2 cos θ – 1) = 0
⇒ sin θ = 23
⇒ θ = nπ + (–1)n 3π , cosθ =
21
⇒ θ = 2nπ ±3π
(C) sin 2θ + cos 2θ + 4 sin θ = 1 + 4 cos θ ⇒ 2 sin θ cos θ + 1 – 2 sin2θ + 4 sin θ = 1+ 4cos θ ⇒ 2 sin θ (cos θ– sin θ) – 4(cosθ– sinθ) = 0 ⇒ (2 sin θ –4) (cos θ – sin θ) = 0 ⇒ sin θ = 2 or sin θ = cos θ ⇒ tan θ = 1 ⇒ θ = nπ + π/4, n ∈ I
(D) cos2 θ = 1/4= cos2 3π
⇒ θ = 2nπ ±3π
11. [A] → r ; [B] → r ; [C] → p ; [D] → q (A)We have b – a = c – b and (c – b)2 = a(b – a) ⇒ (b – a)2 = a(b – a) ⇒ b = 2a and so c = 3a.
Thus a : b : c = 1 : 2 : 3
(B) If the numbers are a and b, then x = 2
ba + and
b = ar3 ⇒ r = 31
ab
Now,
xyz
zy 33 +
=
)ar()ar(xrara2
6333 +
=
x)r1(a 3+
=
2baba
++ = 2
(C) c > 4b –3a ⇒ ar2 + > 4ar –3a ⇒ r2 – 4r + 3 > 0 ⇒ r < 1 or r > 3, But the terms
are positive so r ∈ (0, 1) ∪ (3, ∞)
(D) tan–1
2r21 = tan–1
2r42
= tan–1 )1r2()1r2(1)1r2()1r2(
−++−−+
= tan–1 (2r +1) –tan–1(2r–1)
∴
∑
=
−2
n
1r
1
r21tan (2n+1) – tan–1(1)
= tan–1 (2n + 1) – 4π
XtraEdge for IIT-JEE APRIL 2010 98
∴ tan
∑
=
−2
n
1r
1
r21tan =
1).1n2(11)1n2(
++−+ =
1nn+
∴ ∞→n
lim tan
∑
=
−2
n
1r
1
r21tan =
∞→nlim
1nn+
= 1
Numerical Response type questions : 12. [3] We have
2|z|9 = (2 + cos θ)2 + sin2 θ = 5 + 4 cos θ (1)
and
z3 +
z3 = 4 + 2 cos θ (2)
Eliminating θ from (1) and (2), we get
2|z|9 – 6
+
z1
z1 = – 3
⇒ 3 = 2( z + z) – |z|2 13. [6] Clearly x > 0 and x ≠ 1/5
log5x (5/x) =xlog5logxlog5log
55
55
+−
Putting log5x = t, then equation becomes
t2 +t1t1
+− = 1
⇔ t3 + t2 – 2t = 0 ⇔ t(t – 1) (t + 2) = 0 ⇔ t = 0, 1, – 2 So integral roots are 1 and 5.
14. [2] Let the A.P. be a – d, a, a + d, a + 2d. Note that a
and d must be integers. Also as this is an increasing a + 2d is the largest. We have
a + 2d = (a – d)2 + a2 + (a + d)2 = 3a2 + 2d2 ⇒ 3a2 – a + 2d2 – 2d = 0 As a is real, 1 – 8(d2 – d) ≥ 0
⇒ d2 – d – 81 ≤ 0
⇒ 2
21d
− ≤
83
⇒ 21 –
223
≤ d ≤ 21 +
223
As d is an integer, d = 0, 1 But d ≠ 0, therefore, d = 1. Thus 3a2 – a = 0 ⇒ a = 0 or a = 1/3. As a is an integer, a = 0.
Hence, required number is 2.
15. [1] We have E = 2n + 1C1 + 2n + 1C2 + … + 2n + 1Cn – 2n + 1C0 – 2n + 1C1 – … – 2n + 1Cn [using nCr = nCn – r] = – 2n + 1C0 = – 1. ∴ |E| = 1
16. [3] The given system of equations will have a non-trivial solution if
∆ =t2bac
at2cbcbt2a
++
+= 0
Clearly ∆ is a cubic polynomial in t and has 3 roots.
17. [3] None that it is not given that f is a differentiable function We have
f ′(4) =h
)4(f)h4(flim0h
−+→
= ( )h
)2(f)h4(flim22
0h
−+→
=h
8)h4(lim2/3
0h
−+→
=h
]1)4/h1[(8lim2/3
0h
−+→
=h
1......4h
2318
lim0h
−++
→=
h
.....h838
lim0h
+
→
18. [2]
2x + 2ydxdy = 0 ⇒
dxdy = –
yx
⇒ )3,1(dx
dy = –3
1
Therefore, the equation of the tangent at (1, 3 ) is
y – 3 = (–1/ 3 ) (x – 1)
and the point of intersection of this tangent with the x-axis is (4, 0). The equation of the normal at
(1, 3 ) is y – 3 = 3 (x – 1), and the point of intersection of this normal with the x-axis is (0, 0). Hence the required area is
21 . 4 3 = 2 3
19. [4] The two curves represent parabolas with vertices at (0, 0) and (3, 0). They intersect at (1, 1) and (1, –1), so the required area is
area of OPMQO = 2 (area of OPMO).
XtraEdge for IIT-JEE APRIL 2010 99
O
(1, 1)P
M(3, 0)
Q
Fig.
= 2
−+∫∫
3
1
1
0dx
2x3dxx
= 2
−−
3
1
2/31
0
2/3 )x3(32.
21x
32
= 2 .434
3222
32.
210
32 2/3 =
+=
−−
PHYSICS
1.[D]
d/211′
2′3 3′
2
1 1′3 3′
Cl 2′ 2
Cl
= 2 × d
2)A( 0 ×∈ = 4 .
dA0∈
= 4C
2.[A] 2t = n2
1m2 λ
+
t n mgF2
tmin = n4
λ = 38.14
105.5 7
×× −
= 99.6 nm
3.[D] I = dtdq ; q = it + a; V =
Cq
V = C
ait +
∴ V is proportional to time
4.[A] Phase difference corresponding to y1 = –π/2 and that for y2 = + π/2
∴ Average intensity between y1 and y2
= π1 ∫
π
π−
φ
φ
2/
2/
2max d
2cosI = Imax
π+π
2)2(
Hence required ratio = 21
π+
21
M.C.Q. Type questions :
5.[A,C]
a
l
r
dr
Consider a cylindrical element of radius r, thickness dr. If dR is the resistance of this element then
dR =drr2)r(
πρ l
Total resistance of the cylinder is given by
totalR1 = ∫ dR
1 =lαπ2
∫a
0
3drr
⇒ totalR1 =
lαπ2
4a 4
⇒ totalR1 = 4a
2παl
⇒ ltotalR = 22 )a(
2π
πα
⇒ R = 2A2πα
Since E =l
V (in magnitude)
⇒ E = I
ltotalR (By Ohm's Law)
⇒ E = 2AI2πα
6.[A,D] Consider only one plate as shown in figure.
XtraEdge for IIT-JEE APRIL 2010 100
The field on both sides of plate is shown in figure. Applying Ampere's Circuital Law to the contour C, we get
B
L
C B
2BL = µ0 (jL)
B =2
j0µ
Superimposing, the field due to two plates we get at P both fields cancel each other and at Q, they added to give B0 = µ0j
7.[B,D] v < g(t) ⇒ v < 2g due to Lenz's Law
Also, s <21 gt2
⇒ s < 2g due to Lenz's Law
8.[B,C]
C S
R3
R2
R1
C
R0 R1
V0
V
⇒ R0 =32
32
RRRR
+ and V0 =
32
2
RRVR
+
⇒ τ = C(R1 + R0) = C
+
+32
321 RR
RRR
⇒ q = q0 (1 – e–t/τ) ⇒ q = CV0 (1 – e–t/τ)
⇒ q =32
2
RRCVR
+(1 – e–t/τ)
9.[C] If it was A Y → 2Z Reactant : R = 60 × 8.5 = 510 MeV Product : P = 2 × 30 × 5 = 300 MeV ∆E = – 210 MeV ENDOTHERMIC If it was B W → X + Z R = 120 × 7.5 = 900 MeV P = 90 × 8 + 30 × 5 = 870 MeV ∆E = – 30 MeV ENDOTHERMIC If it was C W → 2Y
R = 120 × 7.5 = 900 MeV P = 2 × 60 × 8.5 = 1020 MeV ∆E = 120 MeV EXOTHERMIC If it was D X → Y + Z R = 90 × 8.0 = 720 MeV P = 60 × 8.5 + 30 × 5.0 = 660 MeV ∆E = – 60 MeV ENDOTHERMIC Column Matching : 10. [A] → p ; [B] → q, r, s; [C] → p,s ; [D] → q,r When cohesive forces are greater then adhesive
forces shape of meniscus is concave from liquid side and pressure is greater in concave side due to surface tension.
11. [A] → r ; [B] → p, s; [C] → q ; [D] → s
(A) T
dTV
dV
P=
⇒
T1
VdV
dT1
P=
(B) λ = ρπ 2d2
kT
(C) Ideal gas law are valid at all temperatures. (D) conceptual Numerical Response type questions : 12. [1]
v1
30º
30ºα
O
α
–v2N
Mv12 v'12
In the figure 12v→
= velocity of ball w.r.t. wedge
before coolision, and 12'v→
= velocity of ball w.r.t. wedge after collision, which must be in vergically upward direction as shown.
In elastic collision 12v→
and 12'v→
will make equal angle (say α) with the normal to the plane. We can show that α = 30º ∴ ∠MON = 30º
Now 2
1
vv = tan 30º =
31
≈ 1
13. [5]
A
h B
Ch
XtraEdge for IIT-JEE APRIL 2010 101
For a cylinder R
T
KK = 2 in case of
pure rolling upto B :
KT =32 mgh
and KR =31 mgh
After B : rotational kinetic energy is constant however translational kinetic energy increases.
At C : KT =32 mgh + mgh =
35 mgh
while KR =31 mgh
∴ R
T
KK = 5
14. [2] According to the formula, the position of the particle is at the centre of the path. It goes by 7 units to the right and then by another 7 units back to the initial position and then it goes to the negative side by 7/2 units for the first time.
So the minimum time is 2 ×4T plus the additional
time (t) to cover 7/2 units on the negative side.
Here ω =2π or
T2π =
2π or T = 4s
∴ tmin = 2 ×44 + t = 2 + t
Now 27 = 7sin
2π t ⇒
2π t = sin 30º =sin
6π
⇒t =31 s
∴ tmin = 2 +31 = 2.33s ≈ 2
15. [6] Let ω be the actual angular velocity of the
satellite from east to west and ωe be the angular speed of the earth (west to east). Then ωrelative = ω – (–we) = ω + ωe ⇒ ω = ωrel – we By the dynamics of circular motion
2RGMm = mω2 R or ω2 = 3
2e
RgR (Q GM =gRe
2)
⇒ ω = 3
2e
RgR ∴ ωrel = 3
2e
RgR + ωe
⇒ ωrel = 213
122
102104.610
××× + 7.27 × 10–5
(Q ωe = 864002π = 7.27 × 10–5)
⇒ ωrel =22.6×10–5 +7.27×10–5 = 30×10–5 rad s–1
∴ τ =relw
2π = 510302
−×π = 2.09×104 s = 5hr 48 min.
16. [1]
V1 =1
4
l8Prη
π , V2 =2
4
l8Prη
π , V3 =3
4
l8Prη
π
and V =l8
Pr 4
ηπ
Now V = V1 + V2 + V3 Substituting the values, we get
l =323121
321
lllllllll
++
=323121
321×+×+×
×× =116 m
17. [1]
At x1 = k3π and x2 = k2
3π
Sin k x1 or sin k x2 is not zero. Therefore, neither of x1 or x2 is a node
∆x = x2 – x1 =
−
31
23
kπ =
k67π
Since k
2π > ∆x >kπ
λ > ∆x >2λ
π
=k
2k
Therefore, φ1 = π
and φ2 = k.∆x =6
7π
∴ 2
1
φφ =
76
≈ 1
18. [1] With an AC source, current in the circuit is maximum when Z = Zmin = R (the resistance of coil)
∴ R =624 = 4Ω
When connected with an dc source of emf 12V.
i =Rr
12+
(r internal resistance of source)
=44
12+
= 1.5 A
19. [1] For incident electron
λ1 = ph =
Km2h =
mVe2h
and shortest wavelength of X-rays is λ2 = Vehc
∴ 2
1
λλ =
c1
me
2V
Substituting the values, we get 2
1
λλ = 1
XtraEdge for IIT-JEE APRIL 2010 102
PHYSICS
1.[C]
2.[B] v ∝ n T∝ because λ = constant
1N4N
++ =
289324 =
1718
17N + 68 = 18N + 18 50 = N
3.[B]
4.[A]
5.[B] u = 0 a = constant v0 = u + at = at
n' =
+vvv 0 n = n +
vtna
6.[C] LC = 1 MSD – 1VSD
LC =
−
30291 MSD =
301 MSD
720 MSD = 360°
1 MSD = 720360° = 1/2°
So LC = 301 ×
21° =
601° =
60'60 = 1'
7.[C] 8.[D] Ticker timer is a better device than a stop watch.
9.[C] M = 2
MM 21 + = 2
62.1230.10 + = 11.46 g
10.[B]
11.[B] YY∆ =
DD2∆ +
l
l∆
YY∆ = 2
4.001.0 +
8.005.0
= 2× 0.025 + 0.0625
YY∆ = 0.05 + 0.0625 = 0.1125
∆Y = 2× 1011 × 0. 1125 = 0.225 × 1011 So (2 ± 0.2) × 1011 N/m2
12.[B]
13.[B]
d
f F
f
It is only possible if object and image coincide.
14.[D] I = I 0 cos2 2φ
2I0 = I 0 cos2
2φ
cos 2φ =
21
φ = 90° = λπ2 ∆x
2π =
λπ2 ∆x
∆x = 4λ ⇒
Ddy =
4λ
y = d4Dλ = 4
7
10141105
−
−
××
×× = 45 × 10–3
y = 1.25 mm
15.[C] →τ = →p × →E = q (2
→a ) ×
→E
Here 2→a = (2 – 1) i + (– 1–0) j + (5 – 4) k
= i – j + k
→E = 0.20 i V/cm = 20 i V/m
∴ →τ = (4 × 10–6) ( )[ ]i20kj–i ×+ = 8 × 10–5
( )jk +
Magnitude of torque τ = 8 2 × 10–5 N-m 16.[D]
– σ
+ σ
+ σ
bc
a
C
B
A
SOLUTION FOR MOCK TEST
PAPER - II AIEEE
XtraEdge for IIT-JEE APRIL 2010 103
Potential of shell A is,
VA = o4
1∈π
σπ+σπ−σπcc4
bb4
aa4 222
= 0∈
σ (a – b + c)
Potential of shell C is,
VC = 04
1∈π
σπ+σπ−σπcc4
cb4
ca4 222
= 0∈
σ
+− c
cb
ca 22
As VA = VC
∴ 0∈
σ (a – b + c) = 0∈
σ
+
− ccb
ca 22
or a – b = ( ) ( )c
baba +− or a + b = c
17.[C]
5
4
3
2
1
B
A
A B
1 2
2 3 5 4
3 4
≡
The capacity of each capacitor is, C0 = d
A0∈
From fig. it is clear that Ceq = 35 C0 =
35
dAo∈
18.[B] Resistivity of conductors increases with increase in temperature because rate of collisions between free electrones and ions increases with increase of temperature. However, the resistivity of semiconductors decreases with increase in temperature because more and more covalent bonds are broken at higher temperatures.
19.[B] let LA and LB be length of parts A and B
Then B
ARR =
B
ALL [as cross-section is same]
Now Lc = 2 LA and (volume)c = (volume)P i.e.Lc × Ac = 2 LA × Ac = LA × AA where Ac = AA are cross-sectional area of part C
and A. ∴ Ac = AA/2
∴ C
ARR =
cALAAL
//
c
A
ρ
ρ =
C
ALL ×
A
CAA
= A
AL2
L × A
AA
2A / = 41
20.[A] Current flowing through potential wire is –
I = rr15
E+
= r16
E
Potential drop across potential wire is,
V = I × 15 r = 16
E15
Potential gradient, K = 600
E1615
∴ 2E = Kl or
2E =
60016E15
× × l
or l = 215
60016×
× = 320 cm.
21.[A]
D i C
O
A BPa
b φ2
φ1φ2 φ1
Q
BAB = BCD = ( )2/b4i0
πµ
(sin φ1 + sin φ1)
= π
µ4
0 . bi4
22 ba
a
+
BBC = BDA = π
µ4
0 .ai4 .
22 ba
b
+
∴ B = BAB + BBC + BCD + BDA
= π
µ4
0 .22 ba
i4
+
+++
ab
ba
ab
ba
= π
µ4
0 . ab
bai8 22 +
22.[A] In tan A position,
π
µ4
0 3dM2 = BH tan 30° =
3BH …….(1)
Magnetic moment of second magnet, M' = (3m)(2 × 2l) = 6M In tan B position,
π
µ4
0 3dM6 = BH tanθ …… (2)
dividing eq. (2) by (1) we get
XtraEdge for IIT-JEE APRIL 2010 104
26 =
3/1tan θ or tan θ = 3 or θ = 60°
23.[B] Magnetic field of solenoid, B1 = l
110 iNµ
Magnet flux of coil, φ2 = N2 B1 A2 = N2
µl
110 iNA2
As φ2 = M i1, so M = 1
2iφ
= l
2210 ANNµ
∴ induced emf, |e| = Mdtdi1
or |e| = l
2210 ANNµ ×
dtid 1
= 30.0
102.13002000104 37– −×××××π × 25.04
= 4.8 × 10–2 Volt
24.[A] Z = ( )2CL2 XXR −+
Here XL = 2πfL = 2 × 3.14 × 500 × (8.1 × 10–3) = 25.4 Ω
and XC = Cf2
1π
= 6105.1250014.321
−××××
= 25.4 Ω
∴ Z = ( ) ( )22 4.254.2510 −+ = 10 Ω
Now irms = Z
Erms = 10
100 = 10 A
∴ VR = irms × R = 10 × 10 = 100 V
25.[D]
26.[C] g
2sinu2 θ = gh2u
( )8.9
452sin)14( 2 =
8.9)h(214 ⇒ h = 10 m
27.[B] Loss in P.E. = gain in K.E.
mg r = 21 mv2 ⇒ v2 = 2g r
ac = r
v2 = 2g
T – mg cos θ = rvm 2
⇒ T = 3 mg
28.[D]
29.[C] Loss in P.E. = gain in K.E. + work done against friction
mg R = 21 m (1.4 gR) + Wf
Wf = 0.3 mgR
Now, W B → C = mg R + 0.3 mgR = 1.3 mgR
30.[C] dtdm × u = M
dtdv
0.5 × 400 = 2000 × 5v∆ ⇒ ∆v = 0.5 ms–1
CHEMISTRY
31.[C] HCO3– can donate a proton to CO3
2– and it can accept a proton to form H2CO3.
32.[C] O F2 ⇒ O = +2 O2 F2 ⇒ O = +1 O2 [PtF6] ⇒ O2
+ + [PtF6]–1
O2+ ⇒ 2x = +1
x = +1/2 33.[A] ∆G = ∆H –T ∆S if ∆H = ⊕ & ∆S = ⊕ & T ∆S > ∆H than ∆G = Θ & process is spontaneous. 34.[C] Calcined Gypsum is calcium sulphate 35.[A] Inter particles forces between CH3COCH3 &
CHCl3 are strong H-bonding. Thus solution shows negative deviation. ∆Vmixing = Negative.
36.[C] Addition of inert gas at constant volume does not cause any effect on the equilibrium.
37.[A] For I order Reaction t1/2 is constant
24tg625.0
24tg25.1
g5.2
24tg5
24tgm10
2/1
2/1
2/1
2/1
=
↓
=
↓
=
↓
=
↓
= 96 hours.
38.[D] 22 Oof.wt
Agof.wtOof.wt.EqAgof.wt.Eq
=
6.1
W8
108=
WAg = 21.6 gm,
XtraEdge for IIT-JEE APRIL 2010 105
39.[B] Solution is decinormal, that is N/10 & x factor is 1, so conc. = 0.1M
[H+] = c.α = 0.1 ×1.3/100 = 13×10–4 pH = –log [13×10–4] = 2.89 40.[A] The no. of atoms in fcc lattice = 4 a = 400 pm = 4×10–10 m = 4×10–8cm
d = 3aNoMn
×× = 38–23 )104(106
604×××
×
d = 6.23 g/cm3.
41.[B] For any Value of l possible values of m are m = –l to + l l = 2, m = –2, –1, 0 +1, +2 So option is (B)
42.[B] C : O : H 6gm : 3.01 × 1023 atoms : 2 mole Ratio ½ : ½ : 2 of mole 1 : 1 : 4 COH4 or CH4O
43.[B]
3 2
1
44.[B] NaNO2/HCl gives HNO2 which gives different products with Pri. and Sec. amines.
45. [A] CH3+ Cl2 ν→h
CH2Cl +HCl
46.[D] 2 CuSO4 + 2 KCN → Cu(CN)2 + K2SO4 2Cu(CN)2 → 2CuCN ↓ + (CN)2 ↑ 3 KCN + CuCN → K3[Cu(CN)4]
47. [B] Blue print process occurs with the help of Iron Compound.
48.[A] Effective nuclear charge increases therefore ionic radius follow the order.
49.[B] xy2 xy + y t = 0 : 600 0 0 teqm : 600-P P P Now : 600 – P + P + P = 800 P = 200 mm Hg
Hgmm100400
200200)Pxy(
)Py)(Pxy(Kp2
=×
==
50.[C] Cell reaction : Zn + Cu+2 → Zn+2 + Cu Cell emf:
Ecell = E°cell – n059.0 log
+
+
2
2
CuZn
So doubling the conc. of ions. Ecell remains same.
51.[A] pKa = –log Ka = –log (1.8×10–5) = 4.7447
[CH3COOH] = L
mol4.05.060
12=
×
[CH3COONa] = 5.0
4.16×82
= 0.4 L
mol
Now, pH = pKa + log
acidsalt
= 4.7447 + log
4.04.0
= 4.7447
52.[B] FeC2O4 → Fe+3 + CO2 +2 +3 +3 +4
Increase = 1
Increase = 1
Total Increase in O.N. = 3 So valence factor of FeC2O4 = 3 KMnO4 →Mn+2 +7 +2 v.f. (KMnO4) = 5 gm E KMnO4 = gm E of FeC2O4 Mole × v.f. = Mole × v.f. 1× 5 = x × 3 ⇒ x = 5/3 53.[A] ∆G = ∆H – T ∆S ∆G = Θ , ∆G < 0 (Spontaneous process) ∆G = ∆H – T ∆S = ∆E + P ∆V – T ∆S (∆G)E,V = 0 + 0 – T ∆S (∆S)E,V = ⊕ ⇒ ∆G = Θ (Spontaneous process)
54.[B]
nCarbocatio2
3CH–HC–3CH
|
3CH|C–3CHH
O2H–3CH–
OH|CH–
3CH|
3CH|C–C3H
°
++→
1, 2-Methyl shift
)stableMore(nCarbocatio3
CH–
3CH|CH–
3CH|C–CHCH–
3CH|C
3CH|C–CH 33
H–33
°
+←= +
55.[C] Factual Q.
XtraEdge for IIT-JEE APRIL 2010 106
56.[D]
OH Zndust→Cl–CH
AlCl
3
3
→ CH3
4KMnO.alk → COOH
57.[D] 58.[A] 4 HCl + O2 → 2 H2O + 2 Cl2 Chlorine is in the form of cloud.
59.[A] Coordination no. = 6 Oxidation no. = 3 no. of d electron = 6 no. of Unpaired d electron = 0
60.[C] Resonance structure should have same number of electron pairs.
MATHEMATICS
61.[A]
P
h
O 60º30º A Let the height of the tower is h 1000 = h cot 30º – h cot 60º
⇒ h = º60cotº30cot
1000−
h = 500 3 m
62.[D] since variance is independent of change in origin. Hence variance of observations 101, 102, ...... 200 is same as variance of 151, 152, .....250.
∴ VA = VB
⇒ B
A
VV = 1
63.[C] centroid of triangle is given by
−++3
3ba,3
3c2
if centroid lie on y axis ⇒ abscissae = 0 c2 + 3 = 0 ⇒ no real c exist if centroid lie on x axis ⇒ ordinate = 0 ⇒ a + b – 3 = 0 ⇒ a + b = 3
64.[C] f(x) = sin x – cos x – kx + b f '(x) = cos x + sin x – k f '(x) = 2 sin(π/4 + x) – k if f(x) is decreases for all x ∴ f '(x) is –ve i. e. k > max. of 2 sin(π/4 + x) i. e. k > 2
65.[A] 0xlim→
−2
x
xxcose
2
= 0xlim→
−+
−22
x
xxcos1
x1e
2
1 + 0xlim→ 2
2
x2/xsin2 = 1 +
21 =
23
66.[A] Second determinant has been obtained from the first by the operation C1 → C1 + 2C2 – 3C3. so its value remains
unchanged
67.[C] given 3sin x – 4sin3x – k = 0 ⇒ 3sin x – 4sin3x = k ⇒ sin3x = k
.........(i) angle A and B satisfy the equation (i) ∴ sin 3A = sin3B = k ⇒ sin3A = sin3B But A > B ⇒ A B Now, sin3A = sin(π – 3B) 3A = π – 3B
A + B = 3π ⇒ C =
32π
68.[B] given statement (p q) ∨ ~ r → (p ∧ r) (F ↔ F) ∨ F → (F ∧ T) T ∨ F → F T → F = F 69.[C]
TFTFFFTTTTTFFFFTFTFFFTTT
p~q)qp(p~qp~qpqp ∧↔∨∧∨
Hence neither tautology nor contradiction 70.[D] Given sequence can be written as
25 ,
1320 ,
910 ,
2320 , ..............
or 820 ,
1320 ,
1820 ,
2320 ..........
XtraEdge for IIT-JEE APRIL 2010 107
which is a H.P. nth term of corresponding A.P.
an = 208 + (n – 1)
205 =
203n5 +
nth term of H.P. = na
1 = 3n5
20+
also 3n5
20+
= 175
5n + 3 = 20 × 5
17 = 68
n = 565 = 13
71.[B] Let any point P divides line joining A(–2, 4, 7) & (3, –5, 8) in ratio λ : 1
then P
−λ+λ
+λλ−
+λλ+−
178,
154,
132
if P lies on plane 2x – k = 0
⇒ x = 2k
2k =
132
+λλ+− given λ = 9
k =
1025 × 2 ⇒ k = 5
72.[C] |a × b| = |a| |b| sin θ
sin θ = 2.5
8 = 54
⇒ cos θ = 53
|a – b|2 = |a|2 + |b|2 – 2a.b = 4 + 25 – 2|a| |b| cos θ
= 29 – 2.2.553 = 29 – 12
|a – b| = 17 73.[C]
1
P2
(0, b)A
B(a, 0)
given length AB = 6 or a2 + b2 = 36
if PBAP =
21
co-ordinates of point P given by
3b2,
3a
locus of point P, a2 + b2 = (3x)2 + 2
2y3
= 36
or 16y
4x 22
+ = 1
b > a then eccentricity e = 2
2
ba1− =
23
74.[D] parabola y = x2 + 2px + p2 + 13 – p2 (y – (13 – p2)) = (x + p)2 vertex is given by (– p, 13 – p2) is 4 units above x-axis ⇒ 13 – p2 = 4 ⇒ p = ± 3 also lies in Ist quadrant ⇒ p < 0 ⇒ p = – 3 75.[C] If lines x2 + 2λx + 2y2 = 0 & (1 + λ) x2 – 8xy + y2 = 0 are equally inclined ⇒ Their bisectors eqn must be same
⇒ eqn
21yx 22
−− =
λxy
& 1)1(
yx 22
−λ+− =
4xy−
are same
⇒ x2 – y2 = λ−
xy & x2 – y2 = λ− /4
xy are same
⇒ – λ = λ−4
⇒ λ2 = 4 ⇒ λ = ± 2 76.[D] given circle in standard form is
x2 + y2 + 2
+
t1t x – 2
−
t1t y + 1 = 0
centre is given by
−−−
t1t,
t1t
or h = –
+
t1t & k = t –
t1
h2 – k2 = 2
t1t
+ –
2
t1t
− = 4
locus x2 – y2 = 4 which is a hyperbola
XtraEdge for IIT-JEE APRIL 2010 108
77.[C] ∫−
4
1
)x(f dx = 4
⇒ ∫−
2
1
)x(f dx + ∫4
2
)x(f dx = 4
⇒ ∫−
2
1
)x(f dx = 4 – ∫4
2
)x(f dx ...........(i)
Q )f(x)– (34
2∫ dx = 7 (given)
⇒ ∫4
2
dx3 – ∫4
2
)x(f dx = 7
⇒ 3[x]42 = 7 + ∫
4
2
)x(f dx
⇒ 3 × 2 = 7 + ∫4
2
)x(f dx ⇒ ∫4
2
)x(f dx = – 1
.....(ii) put this value from (ii) in (i), we get
∫−
2
1
)x(f dx = 4 – (–1) = 5
⇒ ∫−1
2
)x(f dx = – 5
78.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0 (make c1 & c2 +ve) now a1a2 + b1b2 = – 8 + 2 = – 6 < 0 bisector by +ve is acute and contains origin eqn
given by 5
6yx2 +−− = + 52
)7y2x4( +−
– 4x – 2y + 12 = 4x – 2y + 7 8x = 5
79.[C] [ ]∫ −2
1
1)x(gf f 'g(x). g'(x) dx
put f g(x) = t ⇒ f 'g(x) g'(x) dx = dt ∴ required integral [ ]
1
2)x(gflog
log (f g(2)) – log (f g(1)) = 0 Q g(1) = g(2)
80.[C] f(x) = ∫ ++
2
22
x1xsinx sec2x dx
= ∫
+−+
2
22
x1xcos1x sec2x dx
= ∫
+− 2
2
x1xcos1 sec2x dx
= ∫
+− 2
2
x11xsec dx
= tan x – tan–1x + c ∴ f(0) = 0 ⇒ tan0 –tan–10 + c = 0 ⇒ c = 0 ∴ f(x) = tan x – tan–1x ∴ f(1) = tan1 – tan–11 = tan1 – π/4
81.[B] we know x –[x] = x ∴ Domain of x is R & x ∈ [0, 1) but in f(x), x is in denominator & it should
not be equal to zero ∴ x 0 ⇒ x I and domain of sec–1x is R –(–1, 1) ∴ domain of f(x) is R – (–1, 1) – I
82.[C] Q sin–1x is defined for |x| ≤ 1 and sec–1x is defined for |x| ≥ 1 therefore both defined for |x| = 1 ⇒ x = 1,–1 ∴ Df = –1, 1 further f(–1) = sin–1 (–1) + sec–1 (–1) = – π/2 + π = π/2 and f(1) = sin–1(1) + sec–1(1) = π/2 + 0 = π/2 Hence Rf = π/2
83.[B] Let A(z1), B(z2) and C(z3) be the vertices of the triangle then
|z1| = |z2| = |z3| ⇒ |OA| = |OB| = |OC| O being the origin ⇒ O is circumcentre of the triangle, Also, the
triangle is equilateral, therefore circumcentre coincide with the centroid
⇒ origin is centroid
⇒ 3
zzz 321 ++ = 0
⇒ z1 + z2 + z3 = 0 84.[B] Total students n = 100 Average marks x = 72 Total boys n1 = 70 average marks of boys 1x = 75 Total girls n2 = 30
now x = n
xnxn 2111 +
∴ 2x = 2
11
nxnxn −
2x = 30
757072100 ×−× = 65
XtraEdge for IIT-JEE APRIL 2010 109
85.[C] Letters of word 'STATISTICS' are 1A, 2I, 1C, 3S, 3T total = 10 Letters of word 'ASSISTANT' are 2A, 1I, 1N, 3S, 2T total =9 common letters are A, I, S & T
probability of choosing A is = 101 ×
92 =
902
probability of choosing I = 102 ×
91 =
902
probability of choosing S = 103 ×
93 =
909
probability of choosing T = 103 ×
92 =
906
total = 902 +
902 +
909 +
906 =
9019
86.[D] A = – 2, –1, 0, 1, 2 ⇒ n(A) = 5 B = 0, 1, 2, 3 ⇒ n(B) = 4 C = 1, 2 ⇒ n(C) = 2 D = (1, 7),(2, 6),(3, 5),(4, 4),(5, 3)(6, 2),(7, 1), ⇒ n(D) = 7 (A ∪ B ∪ C) = – 2, – 1, 0, 1, 2, 3 ⇒ n(A ∪ B ∪ C ) = 6 n(D) = 7 n(B ∪ C) = 4
87.[B] this is of the form
f '(y)dxdy + f(y). f(x) = Q(x)
put tan y = z
∴ sec2ydxdy =
dxdz
∴dxdz
+ 2xz = x3
∴ ∫pdxe = ∫ xdx2e =
2xe
∴ soln is z 2xe = ∫ 3x x.e
2dx
= 21
∫ 2x x.e2
2x dx
= 21
∫ tdt.et = 21 et (t – 1) + c
∴ tan y 2xe =
21 2xe (x2 – 1) + c
⇒ tan y = c2xe− +
21 (x2 – 1)
88.[D] The given curves are y2 = 8x ...(1) and xy = – 1 ...(2) any tangent to (1) is
y = mx + m2
...(3)
=⇒=∴
+=
=
2a8a4mamxyis
ax42ytogenttanQ
we shall find m so that it touches (2) ∴ from (2) & (3)
1m2mxx −=
+ ⇒ m2x2 + 2x + m = 0 ...(4)
(3) touches (2) if quadratic (4) has equal roots ∴ D = 0 ⇒ 4 – 4m3 = 0 ⇒ m3 = 1 ⇒ m = 1 ∴ required common tangent is y = x + 2
89.[B] h(x) = (f(x))2 + (g(x))2 ∴ h'(x) = 2f(x)f '(x) + 2g(x) g'(x) = 2f(x)g(x) + 2g(x)g'(x) Q f '(x) = g(x) Also f '(x) = g(x) ∴ f"(x) = g'(x) ⇒ – f(x) = g'(x) ∴ h'(x) = 2f(x) g(x) – 2g(x)f(x) h'(x) = 0 ∴ h(x) = constant for all x but h(5) = 11 ∴ h(x) = 1 for all x ∴ h(10) = 11
90.[A] for m1 : Let I & N are assumed single letter in IN order Now total no. of letter are IN T E G E R = 6 letter
these 6 letter can be arranged in row
= 26
× 2 = 720 ways
∴ Letter of words INTEGER can be arranged in
a row = 27
= 2520 ways
Now no. of ways in which IN are not together is m1 = 2520 – 720 = 1800
Now for m2 : I- - - - - R → rest five can arrange
= 25
= 60 ways
∴ 2
1
mm =
601800 = 30
XtraEdge for IIT-JEE APRIL 2010 110
PHYSICS
1.[B] In case of damped vibrations, amplitude decreases exponentially with time
∴ A = A0e–bt or 0A
A = e–bt
or 21 = eb×2 and
0A´A = e–b×b
or 0A´A = (e–2b)3 =
3
21
=
81
2.[C] Here ek
1 = k1 +
k21 +
k41 + ....... ∞
= k1
∞+++ ......
41
21
11
=
− 2/111
k1 =
k2 i.e. ke =
2k
3.[ B] KEmax = 21 kA2 =
21 Mω2A2
8 × 10–3 = 21 × 1 × ω2 × (0.1)2
ω = 4 rad/sec y = A sin(ωt + φ) or y = 0.1 sin (4t + π/4)
4. [B] Here →L = m )vr(
→→× = m v y(–
→k )
(Where y is the vertical distance of particle from x axis)
Here m,v and y all are fixed so →L . remains
constant. 5. [B] Using, weight of floating body = weight of
liquid displaced.
we get V ρ g =
2V (13.6) g –
2V (0.8g)
(buoyant forces of mercury and oil act in opposite direction)
Then, ρ = 2
8.06.13 − = 6.4
6. [C] v´ = s
0
vvvv
−+ v We get
νν´ =
s
0
vvvv
−−
i.e. 89 =
´v340´v340
−+ ∴ (v0 = vs = v´)
i.e. v´ = 20 ms–1
7. [A] Ilog = 1
2
II dB; i.e. 4 = 10 log
)101010(I
492
×× −
Then I2 = 2.5 × 10–4 Wm–2
8. [A] Here work done = pdv and area under the curve given work done
∴ 10 + WCA = 5 or WCA = –5 J
9. [C] For equilibrium F = qE = mg
or neE = 34
πr3ρg or r = 3/1
g4neE3
πρ
10.. [A] i = 111
5.15.15.1++−+ = 0.5 A
As the current has to from A to C to B, for kirchhoff's law, VA = 0.5 × 1 + 1.5 = 1V (Q v = E – ir) VB = 0.5 × 1 + 1.5 = 1V VC = 0.5 × 1 – (–1.5) = 2V
11. [D] R = R1 + (273 – T) α ...(i) or 2R = R0[1 + T´α] ...(ii)
Deciding (i) by (ii) 21 =
α+α
T´1T)-1(273
or 1 + T´α = 2 + (273 – T)2α
or T´ = α
+α 1T)2-(273
12. [A] Current, I = 21 rrR
E2++
P.O. across cell 1 = Ir1 = 21
1
rrREr2
++
For zero p.d. the fall of potential should be equal to in emf.
E = 21
1
rrREr2
++ i.e. R = r1 – r2
SOLUTION FOR MOCK TEST
PAPER - II
BIT-SAT
XtraEdge for IIT-JEE APRIL 2010 111
13. [B] Point P lies on the arms CD and AF so inclusion at P due to them is zero.
Magnetic induction at p due to currents in AB and BC is given by
B1 = B2 = )a2(4
iµ0
πsin 45º
= a28
iµ0
π (Q distance of p from AB or BC is
2a) similarly due to DE and EF
B3 = B4 = a4iµ0
π sin 45º = ⊗
πa24iµ0
Net induction = 2(B1 – B3)
= 2a28
iµ0
π –
a24iµ0
π = –
a8iµ2 0
π
14. [C] v = rω = r × T2π
or T = v
r2π = qB
m2π = q
km
Now mα = 4mp and qα = 2qv
Tp = kαq
mp and Tα = kα
α
qm
= kP
P
q2m4 = 2k
αqmP
⇒ Tα = 2Tp or Tp = 21 Tα
15. [D] i = i0(1 – e–t/τ)
i =
−
−RL/t
e1Rv =
−
−×−
6104.8t
3
e16
12 = 1
(Q i = 1A given) ⇒ t = 0.97 × 10–3 s , i.e. t ≈ 1ms
16. [A] Optical distance between fish and the bird is Differentiating w.r.t.
dtds =
dt´dy +
dtµdy
i.e. 9 = 3 + 34
dtdy
or dtdx = 6 ×
43 = 4.5 ms–1
17. [A] f1 =
−1
µµ
m
g
−
21 R1
R1
f1 =
−1
75.15.1 ×
−
− R1
R1 =
R5.31
i.e. f = 3.5 R. In the medium it behaves as a convergent lens.
18. [D] Reflection of light from plane mirror gives additional path difference of λ/2 between two waves.
∴ Total path difference = 2
3λ + 2π = 2λ
which satisfy the condition of maxima. Resultant intensity ∝ (A2 + A2) [Q I ∝ Α2] 4Α2 = 4Ι
19. [B] Here 1
2
λλ
)()(
20
10
λ−λλ−λ =
12
or 7.34.5
)104.5()105.3(
70
70
−
−
×−λ
×−λ = 12
or λ0 = 11.8 × 10–7 m
But ω = 0
hcλ
=)106.1)(108.11(
)103)(106.6(197
834
−−
−
×××× = 1.05
eV
20. [D] Let whole the energy of electrons be converted in x-rays. eV = hv
or eV = λhc
or λ = eVhc =
)1040)(100.1()103)(106.6(
319
834
××××
−
−
i.e. λ = 3.1 × 10–11 m or λ = 0.31 Å
21. [A] Here 88Ra222 →α 86R218 →−β 87Fr218
→α 85Al214 →−β
80Rn214 →α 84PO210 →α 82Pb206
4α deceys and 2β decays.
22. [A] A1 = 4, A2 = 3 and θ = 2π = 90º
∴ Resultant amplitude,
Α = º90cosAA2AA 2122
21 ++ = 2
221 AA +
= 22 34 + = 25 = 5 unit 23. [C] Using d sin θ = nλ
sin θ = θ = Dλ
∴ Ddy = nλ or y =
dnD
λ
i.e. 3
7
101)106(21
−
−
×××× = 1.2 × 10–3 = 1.2 mm
Distance between first minima on either side of centrar maxima ∆y = 2y = 2.4 mm
24. [B] For constructive inteference
Imax = ( )221 II + = ( )2II2 +
XtraEdge for IIT-JEE APRIL 2010 112
For destuctive interference
Imin = ( )221 II − = ( )2II2 −
Then max
min
II =
2
2
)II2()II2(
+
− = 2
2112
+
− = 341
25. [B] Using decay equation A2 = A1e–λt
or e–λt = 1
2
AA or λt = log
2
1
AA
time t = λ
21 A/Alog = 1.44T loge
2
1
AA
26. [B] Vx = ka
)a4( 2 σπ – b
)b4(k 2 σπ + b
)c4(k 2 σπ
= 4π 04
1πε
σ(a – b + c)
πε
=04
1kQ
= 0ε
σ (a – b + c)
27. [D] Let the given quantity be x1 then,
x = 3π (a2 – b2)h =
3π (a2h – b2h)
= 3π a2h –
3π b2h
Each term has the dimension of x1 then [x] = [a2h] = [L2L] = [L3] and also [x] = [b2h] = [L2L] = [L3] quantity is volume.
28. [B] U = Bx
xA2 +
or A = x
)Bx(U 2 +
Here dimensions of x2 and B should be same. i.e. [B] = [x2] = [L2]
Also [A] =
−
2/1
22
LTML [L2] = [ML7/2T–2]
Then [AB] = [ML7/2T–2] [L2] = [ML11/2T–2
29. [A] v = u
−
´tt1 or
dtdx = u
−
´tt1
integrating, x = u
−
´t2tt
2 + C
at t = 0, n = 0 and c = 0
∴ x = u
−
´t2tt
2 = 10t
−
10´tt
Putting t = 10
x = 10 × 10
−
10101 = 0
30. [A] using h = 21 gt2, we get t1 =
gh2
let t1 be the time taken from instants of jumping to the opening of parachute, then
t1 = 8.9402× = 2.86 sec
His velocity at this point is given by v1
2 = 2gh1 = 2 × 9.8 × 40 = 784 or v1 = 28 ms–1 for the remaining journey, v = v1 + at2
or t2 = a
uv − = 2282
−− = 13sec
∴ total time = t1 + t2 = 2.86 + 13 = 15.86 ≅ 16 s
31. [B] Let u be the velocity of projectile w.r.t. tanks velocity v then
ux = u cos 30 + v; uy = u sin 30º
and T = g
º30sinu2
Range, R1 = uxT = g
º30sinu2 (u cos 30º + v)
for y axis ux
´ = u cos 30º – v and uy´ = u sin 30º
T = g
º30sinu2
Range e, R2 = Tu´x
= g
º30sinu2 (u cos 30º – v)
Then R1 + R2 = gu4 2
(sin 30º cos 30º)
R1 – R2 = gu4 v sin 30º
Eliminating u we get
v2 = 30tan4
g)RR(
)RR(
21
221
+−
= 30tan4
10)200250(
)200250( 2
+− = 24 m2s–2
⇒ 4.9 ms–1
32. [D] Let α be the angle between velocities of pair of particles then relative velocity is given by
vr = α××−+ cosvv2vv 22 = 2 v sin2α
average vr = ∫∫
π
π
α
α2
02
0
d
)2/(sinv2 dα = π4 v
XtraEdge for IIT-JEE APRIL 2010 113
33. [B] For quarter revolution
∆→V =
→
2V – →
1V
angle between →
1V and →
2V is 90º
E
S
N
W
v1
–v1
v2
a
∴ ∆v = 2
122 vv + = 22 vv + = 2 V
Also tan–1
vv = 45º
∴ ∆→v = 2 v south west
34. [D] For vertical motion
H = 21 gt2 or t = g/H2
For horizontal motion, distance covered is given by
2πrn = ut or 2πrn = u g/H2
or u = g/H2r2
4π
35. [A] On descending (Mg – f) – Ma = 0 (where f is the upthrust due to buoyancy) On ascending, f – (M – m)g – (M – m)a = 0
m =
+ gaa2 M
36. [A] The masses will be lifted if the tension of the string is more than the gravitational pull of mass.
50 N20 N
50 N
50 – 2T = 0 or T = 25N So, 5 kg weight cannot be lifted but 2 kg weight
will be lifted
25 – 20 = 2a or a = 25 = 2.5 ms–2
37. [A] On cutting of string QR, the resultant force on m1 remains zero because its weight mg is balanced by the tension is the spring but on block m2 a resultant upward Force (m1 – m2)g is den sped. The block m1 will have no resultant acceleration where as m2 does have an upward
acceleration given by 2
21
mg)mm( −
38. [A] Here v = 2/1
Mpt2
or dtds =
2/1
Mpt2
or ds = 2/1
Mpt2
dt
integrating s = 2/1
Mp2
32 t3/2 + C
at t = 0, S = 0, so c = 0
S = 2/1
M9p8
t3/2
39. [B] Let a small displacement be given to the system
in vertical plane of frame such that ST remains horizontal then let vertical displacement of centres of rods up and QR be y then vertical displacment of centres of VT and RS will be 3y and that of TS will be 4y. Equating total vertical work to zero we get
P Q
W R
T S
W
W
W W
W
W
yyyy
(w + w)δy + (w + w)3 δy + w(4δy) – T(4δy) = 0 or 2w + 6w + 4w = 4T or T = 3w 40.[A] Normal reaction R = mg = 2 × 9.8 N Frictional force, F = µR = 0.2 × 2 × 9.8 = 3.92 N Distance traveled 2 × 5 = 10 m ∴ Work done = f × s = 3.92 × 10 = 39.2 J
XtraEdge for IIT-JEE APRIL 2010 114
CHEMISTRY
1.[D] Fe2O3 + 3CO → 2Fe + 3CO2 160 g 112 g
Pure Fe2O3 in ore = 10
80100× = 80 kg
Iron produced from 80 kg of Fe2O3
= 160112 × 80 = 56 kg
2.[D] ∆v = )x(m4
h∆π
= )1.010)(1011.9)(14.3(4
10626.61031–
34–
×××
−
= 5.79 × 106 ms–1
3.[B] Given n = 0.5
Then
+ 2
2
VanP (V – nb) = nRT
⇒ [P + 2
2
Va)5.0( ] [V – 0.5b] = 0.5 RT
⇒
+ 2V4
aP (2V – b) = RT
4.[B] N2O4(g) 2NO2(g) n0 : 1 0 ne : 1 – 0.5 2 × 0.5 = 0.5 1
Ce : 55.0 M
51
= 0.1 M = 0.2 M
then KC = ]ON[
]NO[
42
22 =
)1.0()2.0( 2
= 1.0
04.0 = 0.4
5.[A] →<<<
strengthAcidicHClOHClOHClOHClO 432
Weakest acid has strongest conjugate base.
6.[C] HCl : NA × VA = (0.4 × 1) × 50 = 20 meq. NaOH : NB × VB = (0.2) × 50 = 10 meq. Q NAVA > NBVB
∴ [H+] = BA
BBAA
VVVNVN
+− =
1001020 −
= 0.1 M = 10–1 M, ∴ pH = 1 7.[D]
(+1) H – O – O – S(x) – O – H
(–1)
(–1) (+1)
(–2)
O
O
(–2)
(–2) 2(+1) + 2 (– 1) + x + 3(–2) = 0 ⇒ x = + 6
8.[C] (+2)(+3) FeC2O4 → Fe3+ + CO2
3+ (+4)
–2e
–1e
∴ Valence factor (FeC2O4) = 1 + 2 = 3
∴ E( FeC2O4) = 3M
9.[D] The fractional of total volume occupied in simple
cube = cubeofvolume
particlesofvolume = 3
3
a2a
34
π
= 6π
10.[A] Cu2+ + e– → Cu+ ; 0
1E = 0.15 V
∴ 01G∆ = –1 × 0.15 × F = – 0.15 F
Cu+ + e– → Cu ; 02E = 0.5 V
∴ 02G∆ = – 1 × 0.5 × F = – 0.5 F
Cu2+ + 2e– → Cu ; 03E = ?
∴ 03G∆ = – 2 × 0
3E × F = – 2F 03E
Also, 03G∆ = 0
1G∆ + 02G∆
= – 0.15 F + (– 0.5 F) = – 0.65 F Now – FE2 0
3 = – 0.65 F
or 03E =
265.0 = 0.325 F
11.[A] 4hrs = 4 half lives 1
→ 2/1t
21
→ 2/1t
41 → 2/1t
81
→ 2/1t
161
fraction left after 4 half lives = 161 or
4
21
fraction reacted in 4 half lives = 1 –161 =
1615
12.[B] A solution showing +ve deviation has higher vapour pressure and lower boiling point.
13.[C] In multi molecular solutions the different layers
hold each other through van der Waal's forces. 14.[A] CS2 + 3O2 → CO2 + 2SO2 ; ∆rH = ? ∆rH° = 0
)P(f HΣ∆ – 0)R(f HΣ∆
= [2(– 297) + (– 393)] – (117) = – 1104 kJ mol–1
15.[C] Aspirin chemically acetyl salicylic acid OCOCH3
COOH
XtraEdge for IIT-JEE APRIL 2010 115
16.[A]
NH2
+ NaNO2 + 2HCl → °− C50
N2Cl
+ 2H2O + NaCl
(CH3)2N H + Cl – N = N →−HCl
(CH3)2N N = N
17.[A] R – C – O – C2H5 + CH3MgBr
O →
:O–
CH3 R
OC2H5
→ −− OHC 52
R – C
O
CH3
→ MgXCH3
XMgO
CH3 R
CH3 →
+H/HOH HO
R
CH3
CH3
18.[C] It is cannizzaro reaction
Cl
CHO
∆ →KOH
Cl
COO–
+
CH2OH
Cl
19.[C] Phenol is less acidic than acetic acid and p-nitrophenol.
20.[A]
OH
+ C2H5I OHHC
.Anhy
52
→
O – C2H5
21.[C] CH ≡ CH 4
42HgSO%1
SOH%40 CH2 = CH – OH
→ − mtautomerisenolKeto CH3 – C – H
O Acetaldehyde
22.[C] The rate of nitration is greater in
hexadeuterobenzene 23.[D] Halogenation on alkene occurs by electrophilic
addition.
24.[A] Twisted boat is chiral as it does not have plane of symmetry.
25.[C]
O C – CH3
Acetophenone has highest dipole moment.
26.[A]
N CH3
CH3CO
N,N-dimethyl cyclopropane carboxamide.
27.[B] NaCl + H2SO4 → NaHSO4 + HCl K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O CrO3 + 2HCl → CrO2Cl2 orange red vapour 28.[C]
Co
Cl
Cl
en
en
+
Cis-d-isomer
Co
Cl
Cl
en
en
+
Cis-l-isomer 29.[B] −
4MnO2 + 16H+ + −42OC →2Mn2+ +2CO2 +
8H2O 30.[C] Lanthanoid contraction takes place. 31.[D] In nitrogen d orbital is absent. 32.[A] HNO3 is acidic in nature. 33.[A] 2KBr + H2SO4 → K2SO4 + 2HBr 34.[A] Due to formation of chelate compound it act as
strong acid and proceed in forward direction. CH – O
BO – CH
CH – O O – CH +H+
35.[B] Na2SO3 + S →NaOH
Sod. thiosulphate Na2S2O3
36.[D] Critical temperature of water is more than O2
due to its dipole moment (Dipole moment of water = 1.84 D, Dipole moment of O2 = 0D.)
37.[D] By the process of zone refining, semiconductors
like Si, Ge and Ga are purified. 38.[C] Half filled orbitals are more stable in
comparison of partial filled. 39.[A] The dipole moment of
CH4 = 0 NF3 = 0.2 D NH3 = 1.47 D H2O = 1.85 D
40.[A] Molecule existence is possible in such case
when no. of bonding electron is greater than antibonding.
XtraEdge for IIT-JEE APRIL 2010 116
MATHEMATICS
1.[D] Clearly all term can Neither be positive nor negative
T1001 = sin log101000 = sin 3 > 0
T10001 = sin log1010000 = sin 4 > 0 ∴ (0 < 2 < π & π < 4 < 2π)
2. [A] sin x < x < tan x in (0, 1) ⇒ tan–1x < x < sin–1 x Altiter : f(x) = sin–1x – x
f '(x) = 2x–1
1 – 1 > 0 ∀ x ∈ (0, 1)
∴ f(x) is increasing function ∴ x > 0 ⇒ f(x) > f(0) = 0 ⇒ sin–1 x > x Similarly g(x) = x – tan–1 x is increasing fun
and x > 0 ⇒ x – tan–1x > 0
3.[A] log2 cos x + log2(1 – tan x) + log2(1 + tan x) – log2 sinx = 1 ⇒ log2(1 – tan2x) – log2 tan x = 1 (cos x > 0, sin x > 0, – 1 < tan x < 1)
⇒ (1 – tan2x) × xtan
1 = 2
tan2x + 2 tan x – 1 = 0 tan x = –1 ± 2
tan x = 2 – 1 (Q 0 < tan x < 1) ⇒ x = π/8 4. [B]
P
A
r
α B
r
α
AB = 2r sin α/2
h = AB tan α = 2r sin 2α tan α
5.[C] Let vertex P be (h, k), then perpendicular distance of P from the base x = a is |h – a|
∴ Since length of the base is 2a, we have
21 × 2a|h – a| = a2
⇒ |h – a| = a (a ≠ 0) So h – a = – a or h – a = a
∴ h = 0 or h = 2a ∴ locus of P is x = 0 or x = 2a
6. [B] Equation of angle bisector of the pair of straight
lines is bayx 22
−− =
hxy−
⇒ hx2 + (a – b)xy – hy2 = 0
⇒ h + (a – b) xy – h
2
xy
= 0
Now, y = mx is one of the bisector ∴ hm2 – (a – b)m – h = 0 h(m2 – 1) = (a – b)m
⇒ m
1m2 − = h
ba −
7.[C] x2 + y2 + 2gx + 2fy + c = 0 passes through all
the four quadrants ⇒ origin in an interior point ⇒ c < 0 8. [A] two normal are x – 1 = 0 and y – 2 = 0, their
point of intersection (1, 2) is the centre & radius of circle perpendicular distance from centre (1, 2) on tangent 3x + 4y = 6
= 169
62.43+
−+ = 1
∴ equation of circle is (x – 1)2 + (y – 2)2 = 12 ⇒ x2 + y2 – 2x – 4y + 4 = 0
9.[B] x2 – 4x + 6y + 10 = 0 ⇒ (x – 2)2 = –6(y + 1) tangent to the vertex is y + 1 = 0 circle drawn on focal distance as diameter
always touch the tangent at the vertex i.e. the line y + 1 = 0.
10. [B] Given ellipse is 25x2
+ 9y2
= 1
a2 = 25, b2 = 9, e = a
ba 22 − = 54
⇒ ae = 4 ∴ Foci of ellipse are (± ae, 0) = (± 4, 0) For hyperbola e = 2 ⇒ 2a = 4 ⇒ a = 2 Also b2 = a2(e2 – 1) ⇒ b2 = 4 × 3 = 12 ∴ equation of hyperbola
4
x2 –
12y2
= 1
⇒ 3x2 – y2 – 12 = 0
XtraEdge for IIT-JEE APRIL 2010 117
11.[B] ∴ →α &
→β are two mutually perpendicular
unit vector.
∴ →α ×
→β is a unit vector perpendicular to
both →α &
→β . So we can consider
→α ,
→β ,
→α ×
→β as i , j & k .
Given vector are coplanar so
bcc101caa
= 0
⇒ a(– c) + a(c – b) + c2 = 0 ⇒ c2 = ab
12. [D] S1 = x2 + y2 + z2 + 2x – 4y – 4z – 7 = 0 centre C1 = (–1, 2, 2) and radius r1 = 4 S2 = x2 + y2 + z2 + 2x – 4y – 16z + 65 = 0 centre C2 = (–1, 2, 8) radius r2 = 2 C1C2 = 6; r1 + r2 = 6 ∴ sphere S1 & S2 touches externaly ∴ point of contact divides C1C2 in the ratio 4 : 2 ∴ point of contact = (–1, 2, 6)
13.[D] y = f(x) = 3 – 8x4x
42 +−
⇒ (3 – y)x2 – 4(3 – y)x + 20 – 8y = 0 x ∈ R ∴ D ≥ 0 ⇒ 16(3 – y)2 – 4(3 – y) (20 – 8y) ≥ 0, y ≠ 3 ⇒ –y2 + 5y – 6 ≥ 0; y ≠ 3 ⇒ (y – 2) (y – 3) ≤ 0 ⇒ 2 ≤ y < 3
14. [B] x > 0, g(x) is bounded
∴ nx
nx
n e1)x(ge)x(flim
++
∞→
nx
nx
n e11e/)x(g)x(f
lim+
+∞→
= f(x)
[g(x) is bounded ⇒ nxe)x(g ⇒
infintefinite = 0]
15.[B] a1 = 1, an = n(1 + an–1)
⇒ 1 + an–1 = n
a n
∴ ∞→n
lim
+
1a11
+
2a11 ....
+
na11
∞→n
lim
+
1
1
aa1
+
2
2
aa1 ....
+
n
n
aa1
=∞→n
lim2
a 2 .3
a3 ........ 1n1a n
++ .
n21 a...aa1
= 1n
alim 1nn +
+
∞→ =
1n)n1)(a1(lim n
n +++
∞→
=
+
∞→ na
n1lim n
n=
−+
−+ −
∞→ 1na
1n1
n1lim 1n
n
=
+++−
+∞→ 1
a11
21....
1n1
n1lim 1
n
=
+++∞→ n
1....2
11
11limn
Q a1 = 1
= e
16.[A] 2 sgn 2x =
>=<−
0x;20x;00x;2
f(x) = |2 sgn 2x| + 2 =
=≠
0x20x4
∴ By defining f(0) = 4, f(x) will become continuous function at x = 0 as then
f(0 – 0) = f(0 + 0) = 4 Hence at x = 0 f(x) has removable discontinuity 17.[A] Q 1 < x < 2 ⇒ [x] = 1
∴ f(x) = cos
−
π 3x2
= sin x3
f ´(x) = cos x3 . 3x2
⇒ f ´
π32
= 3 cos2π .
3/2
2
π = 0
18.[C] yex = cos x ...(1) yex + y1ex = – sin x ...(2) again differentiating yex + y1ex + y1ex + y2ex = – cos x yex + 2y1ex + y2ex = – yex (from (1)) ⇒ 2y + 2y1 + y2 = 0 ...(3) again differentiating 2y1 + 2y2 + y3 = 0 again differentiating 2y2 + 2y3 + y4 = 0 ...(4) from (1) (2) & (3) 4y + y4 = 0
∴ y
y4 = – 4
19.[A] fog = I ⇒ fog(x) = I(x) = x ∴ fg(x) = x ⇒ f´g(x) × g´(x) = 1 ⇒ f´g(a) × g´(a) = 1 ⇒ f´(b) × 2 = 1
⇒ f´(b) = 21
XtraEdge for IIT-JEE APRIL 2010 118
20.[C] Let P(h, k) be one of point of contact then k = sin h ...(1) equation of tangent is y – k = cos h (x – h) which parries through origin ∴ k = h cos h ...(2) from (1) & (2)
2
2
hk + k2 = sin2h + cos2h = 1
⇒ k2 + k2h2 = h2 ⇒ x2 – y2 = x2y2 ∴ locus of (h, k) is x2 – y2 = x2y2 21.[B] x + y = 16, x, y > 0 s = x3 + y3 = x3 + (16 – x)3
dxds = 3x2 – 3(16 – x)2
dxds = 0
⇒ x2 = 256 – 32x + x2 ⇒ x = 8
2
2
dxsd = 6x – 6(16 – x) = 6[2x – 16]
at x = 8
2
2
dxsd = 0 and 3
3
dxsd = 12 ≠ 0
Hence there in no minimum exist.
22.[C]
a1
b1
a1
b1
22
−
− =
2
3
c1c2
−
− =
c2
⇒ b1 +
a1 =
c2
⇒ c = ca
ab2+
∴ c is H.M. of a & b
23.[B] I = ∫ + xcos1 2 sin 2x cos 2x dx
put 1 + cos2x = t2 ⇒ –2sin x cos x dx = 2tdt ⇒ – sin 2x dx = 2tdt
∴ I = – ∫ − )1xcos2).(dtt2.(t 22
= – ∫ − dt)3t2(.t2.t 2
= –2
−
3t3
5t2 35
+ c
= – 54 (1 + cos2x)5/2 + 2(1 + cos2x)3/2 + c
= (1 + cos2x)3/2
++− 2)xcos1(
54 2 + c
= 52 (1 + cos2x)3/2 (3 – 2cos2x) + c
24.[A] 1 – 3
x2 +
4x4
– .......... = cos x
∴ I = ∫ dxxcos = sin x
25.[A]
∫
∫
π
π
−
π→
π
−
−
2
2
x
4/
x
2/
tcos
2x dt
2t
dt)12(lim
00
x2.
2x
12limxcos
2x
π
−
−−
π→
= π−
−
π→ x4
xsin.2n2limxcos
2x
l = π
2nl
26.[A] y = 222 n
6
2
2n
4
2
2n
2
2n n31
n21
n11lim
+
+
+
∞→
....2nn2
2
2
nn1
+
log y = ∑=
∞→
+
n
1r2
2
n nr1loglim . 2n
r2
= ∑=
∞→
+
n
1r2
2
n nr1loglim .
nr2 .
n1
= ∫ +1
0
2 dx)x1log(x2
= ∫2
1
dttlog = ( )21ttlogt −
= 2 log 2 – 1 = loge4
⇒ y = 4/e 27.[B] required area
–1
0
1 y = 2x2
1/√2
XtraEdge for IIT-JEE APRIL 2010 119
2
1 – ∫2/1
0
2dxx2
⇒ 2
1 –32 2/1
03 ]x[
⇒ 2
1 – 32 .
221 =
232 =
622
28.[B] ´´y = (y´ + 3)1/3 ⇒ (y´´)3 = (y´ + 3)2
⇒ 3
2
2
dxyd
–
2
dxdy
– 6
dxdy – 9 = 0
∴ order is 2 & degree = 3 29.[A] x18 = y21 = z28 ⇒ 18 log x = 21 log y = 28 log z ⇒ logy x = 7/6, logz y = 4/3, logxz = 9/14 Now, 3, 3 logyx, 3 logz7, 7 logx2
= 3, 27 , 3 ×
34 , 7 ×
149
= 3, 27 , 4,
29
which are in A.P. 30.[B] log2x + log2y ≥ 6 Here x > 0, y > 0 ∴ log2xy ≥ 6 ⇒ xy ≥ 26 ⇒ xy ≥ 64 Now, A.M ≥ G.M.
∴ 2
yx + ≥ (xy)1/2
⇒ x + y ≥ 2(64)1/2 x + y ≥ 2 × 8 ⇒ x + y ≥ 16 ∴ (x + y)min = 16 31.[A] f(x) = x3 + x2 + 10x + sinx f´(x) = 3x2 + 2x + 10 + cos x
= 3 2
31x
+ +
329 + cos x > 0 ∀ x
⇒ f(x) is strictly increasing Also x → ∞ ⇒ f(x) → ∞, x → – ∞ ⇒ f(x) → – ∞ ∴ f(x) has only one real root. 32.[B] Let roots be (2k – 1) & (2k + 1) k ∈ N
the Sum of roots : 4k = – ab
Q a ∈ R+, b < 0 as k ≥ 1
We have – b = 4ak ⇒ – b ≥ 4a ⇒ |b| ≥ 4a b < 0 ∴ |b| = –b 33.[B] Let G.P. be a + ar + ar2 ...... G.P is infinite so – 1 < r < 1 G.P. is decreasing ⇒ r > 0 so 0 < r < 1 and therefore a > 0 f´(x) = 3x2 + 3 > 0 ⇒ f(x) is strictly increasing function ∴ f(x)max on [–2, 3] is f(3) = 27 & f´(0) = 3
∴ r1
a−
= 27 & a – ar = 3
⇒ r = 32 or
34 Q r < 1
∴ r = 32 & if r =
32 ; a = 9
∴ Sum of first three terms = 9 + 6 +4 = 19
34.[B] If z1, z2 & z3 vertex of equilateral triangle then 2
1z + 22z + 2
3z = z1z2 + z2z3 + z3z1 ∴ (a + i)2 + (1 + ib)2 + 0 = (a + i) (1 + ib) + 0 +
0 ⇒ a2 – 1 + 2ia + 1 – b2 + 2ib = a + iba + i – b ⇒ a2 – b2 + i(2a + 2b) = a – b + i(ab + 1) on comparing a2 – b2 = a – b and 2a + 2b = ab + 1 ⇒ (a – b) (a + b – 1) = 0 & 2a + 2b = ab + 1 ⇒ a = b or a + b = 1 ....(1) ⇒ 2a + 2b = ab + 1 .... (2) Now from (1) take a = b put in (2) 2a + 2a = a2 + 1 ⇒ a2 – 4a + 1 = 0 ⇒ a = 2 ± 3
Q a < 1 ⇒ a = 2 – 3
Q a = 2 – 3 & a = b = 2 – 3 It we take a + b = 1 & put in (2) then it becomes
ab = 0 which not possible because a & b lies between 0 and 1
35.[A] S =
+
+
+−∑
=
....87
43
21)1(
rr
r
n
0r
r nCr
= rn
r
n
0r
r C.21)1(∑
=
− + rn
rn
0r
r C.43.)1(
−∑
=
+ rn
rn
0r
r C.87.)1(
−∑
=
+ ....
= n
211
− +
n
431
− +
n
871
− + .....
= n21 + n22
1 + n321 + ..... =
121
n −
XtraEdge for IIT-JEE APRIL 2010 120
36.[B] The general term in the expansion of (x1 + x2 +
...xn)n given ..... n21 p...pp
n. 1p
1x 2p2x .... mp
mx ,
p1 + p2 + p3 .... + pm = n Now in (1 + x + y – z)9, coefficient of x3y4z = coeft of u0x3.y4z1 in (u + x + y – z)9
= 1430
9 × (–1)1 = –2 . 9C2 . 7C3
37.[B] x1
ex
− = B0 + B1x + B2x2 + .....
⇒ ex = B0 – B0x + B1x – B1x2 + B2x2 – B2x2 + ....
⇒ ex = B0 + (B1 – B0)x + (B2 – B1)x2 + .....
⇒ 1 + x + 2
x2+
3x3
+ 4
x4+ ....
= B0 + (B1 – B0)x + (B2 – B1)x2 + ... ∴ Bn – Bn –1 is coeff. of xn
On comparing coeff. of xn = n1
38.[A] x + y + z + 12 = 0 x, y, z are negative integers Let x = – a, y = –b, z = – c, a, b, c are +ve integer then required number of
points (x, y, z) = Number of positive integral solution of a + b + c = 12 = 12–1C3–1 = 11C2 = 55
39.[A] p1 = 12
662 p2 =
11
65
2
1
pp = 11
6512
662× = 1
40.[C] 2f(x) = )x/1(f1
)x(f)x/1(f)x(f −
= f(x) .
x1 – f
x1 + f(x)
⇒ f(x) + f
x1 = f(x) . f
x1
⇒ f(x) = 1 ± xn f(2) = 17 ⇒ 1 ± 2n = 17 ⇒ ± 24 = 16 ∴ +ve sign will be take ⇒ 2n = 16 ⇒ n = 4 Now, ∴ f(x) = 1 + x4 ⇒ f(5) = 54 + 1 = 626
41.[C] A is idenpotent ⇒ A2 = A
A2 =
20x1
20x1
=
40x31
≠ A
∴ not possible for any x 42.[A] for any a ∈z ⇒ a = 20a ⇒ a R a ∀ a ∈ z ∴R is reflexive a R b ⇒ a = 2kb, k ∈ z ⇒ b = a.2–k, – k ∈ z ⇒ b R a ∴ R is symmetric Let a R b, b R c ⇒ a = b2 1k , b = c2 2k a = 1k2 c2 2k = c2 21 kk + , k1 + k2 ∈ z ⇒ a R c ∴ R is transitive Hence R is an equivalence Relation. 43.[D] Q A.m ≥ G.m
∴
+
ab
ba
21 ≥
ba.
ab = 1 ⇒
ba +
ab ≥ 2
Similarly cb +
bc ≥ 2 &
ca +
ac ≥ 2
Adding we get
ba +
ab +
cb +
bc +
ca +
ac ≥ 6
⇒ a
cb + + b
ac + + c
ba + ≥ 6
∴ minimum value is 6.
44.[A] f(x) = x1 ⇒ f ´(x) = – 2x
1
∴ ab
)a(f)b(f−− = f ´(x1)
⇒ b1 –
a1 = (b – a)
− 2
1x1 ;
a < x1 < b ⇒ x1
2 = ab ⇒ x1 = ab
45.[C] ∫ xcot4 dx = ∫ xcot2 (cosec2 x –1)dx
= ∫ xcot2 cosec2x dx – ∫ − )1xec(cos 2 dx
= – 31 cot3x + cotx + x + c
∴f(x) = – 31 cot3x + cotx + x + c +
31 cot3x – cotx
= x + c
XtraEdge for IIT-JEE APRIL 2010 121
∴ f
2x =
2π + c ⇒
2π =
2π + c
⇒ c = 0 ∴ f(x) = x
LOGICAL REASONING 1.[B] The given sequence is a combination of two
series: I series : 11, 17, 23, (?) II series : 12, 18, 24 Pattern in both is + 6 So, missing number = 23 + 6 = 29
2.[B] Dum Dum is an airport in calcutta and Palam is an airport in Delhi.
3. [D] All except mustard are food grains, while
mustarel is an oilseed. 4.[C] 5. [B]
6.[D]
7.[C] The 3rd figure in each row comprises of parts which are not common to the first two figures.
8. [C]
9.[D]
10.[B] The number of each type of figures decrease by 1 at each step from left to eight in each row.
ENGLISH 1.[B] Assure : (persuade that all is well) Hence, irrelevant meaning. Ensure : (guarantee) This is a relevant option as it properly suits to
the meaningful expression. Insure : (to cover against any loss) Hence, irrelevant meaning.
Accept : Irrelevant meaning in respect of the sentence. 2.[B] burst : (punctured) Irrelevant meaning. bust : (collapsed) Quite relevant meaning. Hence, correct option. bursted : irrelevant because this is an improper
form of the verb 'burst'. busted : Irrelevant because this is an incorrect form of
the verb 'bust' 3.[A] deduce : (to conclude) (to infer) This word really suits to the given sentence
making it a meaningful one. deduct : (to take away) Irrelevant, because it means something else. e.g. Tax is 'deducted' from his salary every year. reduce : (to decrease) Different meaning makes the sentence
meaningless. Hence, this is an incorrect option. Conduce : (to suit) Irrelevant word.
4.[C] Sky (Firmament) Irrelevant 'opposite' Firmament : irrelevant 'opposite'. nadir : (lowest point) 'Zenith' means highest point'. Hence this is the most suitable word in opposite, naive (Simple) Irrelevant word that doesn't serve any purpose.
5.[A] Hungry : (a voracious eater) (a voracious reader)
Wild : (Irrelevant) It's Synonym is 'Savage'. Quick : (Soon) therefore, irrelevant. Angry : (furious) Hence, irrelevant.
6.[C] Journey man : (Irrelevant) Because it is a person who journeys regularly on
a particular route. Tramp : (Irrelevant) because it means 'a vagabond' who gets about
purposelessly. Itinerant : (Relevant) It's a person who moves from one place to
another during his travel. Mendicant : (Irrelevant) It's a religious preacher who goes from place to
place in the form of a beggar.
7.[B] To dislocate : irrelevant meaning. To lose one's temper : Quite Relevant It's often used when a person is about to get
angry.
XtraEdge for IIT-JEE APRIL 2010 122
To take off : irrelevant To be indifferent : Irrelevant 8.[A] took to : (to be accustomed to/to be addicted to) Correct because it suits to the sentence when
Gandhi Ji was addicted to smoking. took for : (to be mistaken while recognising) irrelevant took in : (to deceive someone) irrelevant took up : (to adopt) irrelevant 9.[A] Get someone to break the box. Correct answer because : • The given sentence is in Passive Voice which requires its answer in Active Voice. • The given verb is Causatives in Imperative. • This option is Causative Active Voice in imperative form. They have broken the box. Incorrect Answer : Because • The given verb is not causative. • The given sentence is not imperative. Have the broken box. Incorrect : Because • The verb is not causatives. • 'Broken' has been used as an adjective in this
option. Break the box Incorrect because : • The subject 'you' (implied) wan't break it but
will get someone to break it.
10.[B] He asked how shabby I was looking. (Incorrect option) because : The required answer (type of sentence) is
wrong. He exclaimed with disgust that I was looking
very shabby. (Correct answer because) • This option is exclamatory. • Past Indefinite Tense has been used. • The mood of the speaker is correct. He exclaimed with sorrow that they were
looking much shabby. (Incorrect option because) : • Mood of exclamatory sentence is wrong. • 'much' will be replaced with 'very' • 'They' won't be used as a singular subject is
required. He told that I was looking much shabby. Incorrect answer because – • Type of sentence is assertive whereas the
required type is 'exclamatory'. • 'much' is to be replaced with 'very'.
11.[C] Neigh : Correct spelling as it means 'the cry of horse'.
Reign : Correct spelling as it means 'the controlling chord of an animal.'
Niece : Incorrect spelling as the correct one is 'niece'. (Opposite of 'Nephew').
Neither : Correct spelling. It is a conjunction to be used with 'nor' for one of two options.
12.[C] I wonder : No error in it. What he has done with the book. No error I lend him (Erroneous) because there is an error of
'Tenses'. The word 'lend' is to be 'lent'. No error : There is an error. 13.[A] Distraught, awry : Correct answer : 'Distraught' means to 'get
upset' and 'awry' means in 'disorder'. Frustrated, Magnificently : Both are opposite. One is positive and the other
one is negative. Therefore, no meaningful sentence.
Elated, Wild : No co-ordination, therefore incorrect answer. Dejected, splendidly : No co-ordination, therefore incorrect answer. 14.[D] Interesting : " . . . and only a few were . . . . " phrase shows
that something opposite is required here. The given option is not opposite to 'trivial'. Hence, Irrelevant option.
Practical : Like aforesaid logic, this option also is
irrelevant. Complex : 'Complex' can't be opposite to 'trivial'.
Therefore, can't be relevant. Significant : This is the relevant option making the sentence
quite meaningful with two contradictory words, i.e. trivial and 'Significant connected with the phrase ". . . . . and only a few were . . . . . "
15.[A] Paths, grave : It is a meaningful pair of words to make the
sentence idiomatically correct. Ways, happiness : Not a meaningful pair. Hence, irrelevant option. Acts, prosperity : Not a meaningful pair. Hence, irrelevant option. Achievements, Suffering : Not a meaningful pair. Hence, irrelevant option. Hence, inappropriate
XtraEdge for IIT-JEE APRIL 2010 123
XtraEdge for IIT-JEE APRIL 2010 124
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